Q.Integrate the following function: $\frac{1}{x^2 - 9}$

Showing the 12 most recent of 33 on this concept.

1. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.The partial fraction decomposition of $\dfrac{9x-7}{(x+3)(x^2+1)}$ is (A) $\dfrac{17}{5(x+3)} - \dfrac{(17x-6)}{5(x^2+1)}$ (B) $\dfrac{-17}{5(x+3)} - \dfrac{(17x-6)}{5(x^2+1)}$ (C) $\dfrac{17}{5(x+3)} + \dfrac{(17x-6)}{5(x^2+1)}$ (D) $\dfrac{-17}{5(x+3)} + \dfrac{(17x-6)}{5(x^2+1)}$

   ›Reveal solutionSolution

   The key idea is to decompose a rational function with an irreducible quadratic factor into a sum of a linear-over-quadratic term and a constant-over-linear term, then solve for the unknown coefficients. The correct decomposition is option (D).

   When you see a denominator with a linear factor $(x+3)$ and an irreducible quadratic factor $(x^2+1)$, the standard partial fraction form is:

   $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$

   The numerator over the quadratic must be linear (degree 1) because the denominator is degree 2 — a common point where students mistakenly put just a constant. The goal is to find $A$, $B$, and $C$ by clearing denominators and equating coefficients.

   1. Set up the equation Multiply both sides by $(x+3)(x^2+1)$:

   $9x - 7 = A(x^2+1) + (Bx+C)(x+3)$

   2. Expand the right-hand side

   $A(x^2+1) = Ax^2 + A$

   $(Bx+C)(x+3) = Bx^2 + 3Bx + Cx + 3C = Bx^2 + (3B+C)x + 3C$

   Adding them:

   $9x - 7 = (A+B)x^2 + (3B+C)x + (A + 3C)$

   3. Equate coefficients Comparing coefficients of $x^2$, $x$, and the constant term gives three equations:

   $\begin{cases} A + B = 0 & \text{(coefficient of }x^2\text{)} \\ 3B + C = 9 & \text{(coefficient of }x\text{)} \\ A + 3C = -7 & \text{(constant term)} \end{cases}$

   4. Solve the system From the first equation: $B = -A$. Substitute into the second: $3(-A) + C = 9 \implies -3A + C = 9$. The third equation is $A + 3C = -7$. Solve these two:

   $\begin{aligned} -3A + C &= 9 \\ A + 3C &= -7 \end{aligned}$

   Multiply the first equation by 3: $-9A + 3C = 27$. Subtract the second equation from this:

   $(-9A + 3C) - (A + 3C) = 27 - (-7) \implies -10A = 34 \implies A = -\frac{34}{10} = -\frac{17}{5}$

   Then $B = -A = \frac{17}{5}$.

   Substitute $A$ into $A + 3C = -7$:

   $-\frac{17}{5} + 3C = -7 \implies 3C = -7 + \frac{17}{5} = -\frac{35}{5} + \frac{17}{5} = -\frac{18}{5} \implies C = -\frac{6}{5}$

   5. Write the decomposition With $A = -\frac{17}{5}$, $B = \frac{17}{5}$, $C = -\frac{6}{5}$, we have:

   $\frac{9x-7}{(x+3)(x^2+1)} = \frac{-\frac{17}{5}}{x+3} + \frac{\frac{17}{5}x - \frac{6}{5}}{x^2+1}$

   Factor out $\frac{1}{5}$:

   $= -\frac{17}{5(x+3)} + \frac{17x - 6}{5(x^2+1)}$

   Watch out

   A common mistake is to forget the minus sign on the first term or to put a minus sign in the numerator of the second term. Check by combining the right-hand side back — it must give the original numerator $9x-7$.

   ✓Final answer

   The correct option is (D): $\displaystyle -\frac{17}{5(x+3)} + \frac{17x-6}{5(x^2+1)}$.

2. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.$\int \frac{1}{(x-2)(x^2+1)} \, dx =$ (A) $\log \frac{\sqrt{x^2+1}}{|x-2|} + 2 \tan^{-1} x + c$ (B) $\log \frac{|x-2|}{x^2+1} + 2 \tan^{-1} x + c$ (C) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{7+x^2}} + 2 \tan^{-1} x \right] + c$ (D) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{1+x^2}} - 2 \tan^{-1} x \right] + c$

   ›Reveal solutionSolution

   The integral is solved by partial fraction decomposition, leading to a combination of a logarithmic term and an arctangent term. The correct result is $\frac{1}{5}\left[\log\frac{|x-2|}{\sqrt{x^2+1}} + 2\tan^{-1}x\right] + c$, which matches option (C).

   We have a rational function whose denominator factors into a linear term $(x-2)$ and an irreducible quadratic $(x^2+1)$. The natural strategy is partial fractions, because the integrand is a proper rational function (degree of numerator < degree of denominator). This will break the integral into simpler pieces we can integrate directly.

   ---

   1. Set up the partial fraction decomposition

   We write:

   $\frac{1}{(x-2)(x^2+1)} = \frac{A}{x-2} + \frac{Bx + C}{x^2+1}$

   The numerator for the quadratic term is linear because $x^2+1$ cannot be factored further over the reals.

   2. Clear denominators

   Multiply both sides by $(x-2)(x^2+1)$:

   $1 = A(x^2+1) + (Bx + C)(x-2)$

   3. Expand and collect like terms

   $1 = A x^2 + A + Bx(x-2) + C(x-2)$

   $= A x^2 + A + Bx^2 - 2Bx + Cx - 2C$

   $= (A + B)x^2 + (-2B + C)x + (A - 2C)$

   4. Equate coefficients

   Since the left side is $1 = 0x^2 + 0x + 1$, we have:

   $\begin{cases} A + B = 0 \\ -2B + C = 0 \\ A - 2C = 1 \end{cases}$

   5. Solve the system

   From the first equation: $B = -A$.

   From the second: $C = 2B = -2A$.

   Substitute into the third: $A - 2(-2A) = A + 4A = 5A = 1$, so $A = \frac{1}{5}$.

   Then $B = -\frac{1}{5}$, $C = -\frac{2}{5}$.

   6. Write the decomposed integrand

   $\frac{1}{(x-2)(x^2+1)} = \frac{1/5}{x-2} + \frac{-\frac{1}{5}x - \frac{2}{5}}{x^2+1}$

   7. Integrate term by term

   $\int \frac{1}{(x-2)(x^2+1)} \, dx = \frac{1}{5} \int \frac{dx}{x-2} - \frac{1}{5} \int \frac{x}{x^2+1} \, dx - \frac{2}{5} \int \frac{dx}{x^2+1}$

   • First term: $\int \frac{dx}{x-2} = \log|x-2| + c_1$
   • Second term: Let $u = x^2+1$, $du = 2x\,dx$, so $\int \frac{x}{x^2+1} dx = \frac{1}{2} \log(x^2+1) + c_2$
   • Third term: $\int \frac{dx}{x^2+1} = \tan^{-1} x + c_3$

   8. Combine results

   $= \frac{1}{5} \log|x-2| - \frac{1}{5} \cdot \frac{1}{2} \log(x^2+1) - \frac{2}{5} \tan^{-1} x + C$

   $= \frac{1}{5} \left[ \log|x-2| - \frac{1}{2} \log(x^2+1) - 2 \tan^{-1} x \right] + C$

   9. Simplify the logarithmic part

   Recall $-\frac{1}{2}\log(x^2+1) = \log\left( (x^2+1)^{-1/2} \right) = \log\frac{1}{\sqrt{x^2+1}}$.

   So:

   $\log|x-2| + \log\frac{1}{\sqrt{x^2+1}} = \log\frac{|x-2|}{\sqrt{x^2+1}}$

   Thus:

   $\int \frac{1}{(x-2)(x^2+1)} \, dx = \frac{1}{5} \left[ \log\frac{|x-2|}{\sqrt{x^2+1}} - 2 \tan^{-1} x \right] + C$

   Watch out

   Option (D) has a minus sign before $2\tan^{-1}x$, but our result has a minus sign as well — wait, check carefully: Our result is $\frac{1}{5}[\log\frac{|x-2|}{\sqrt{x^2+1}} - 2\tan^{-1}x]$, which is exactly option (D)? No, look again: Option (D) says $-2\tan^{-1}x$, but option (C) says $+2\tan^{-1}x$. Did we get a sign error? Let's verify the sign of the arctangent term.

   We had $-\frac{2}{5}\int \frac{dx}{x^2+1} = -\frac{2}{5}\tan^{-1}x$. So indeed the coefficient is negative. That matches option (D), not (C). But wait — let's re-check the original options:

   (A) $\log \frac{\sqrt{x^2+1}}{|x-2|} + 2 \tan^{-1} x + c$

   (B) $\log \frac{|x-2|}{x^2+1} + 2 \tan^{-1} x + c$

   (C) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{7+x^2}} + 2 \tan^{-1} x \right] + c$

   (D) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{1+x^2}} - 2 \tan^{-1} x \right] + c$

   Our result: $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{x^2+1}} - 2 \tan^{-1} x \right] + c$ matches (D) exactly. But option (C) has $\sqrt{7+x^2}$ (a typo? likely meant $\sqrt{1+x^2}$) and a plus sign. So the correct match is (D).

   Tip

   Always double-check the sign of the arctangent term: the partial fraction gave $-\frac{2}{5}\tan^{-1}x$, so the minus sign is correct.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

3. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.The partial fraction decomposition of $\dfrac{3x+1}{(x-1)^2(x+2)}$ is (A) $\dfrac{4}{3}\dfrac{1}{(x-1)^2} + \dfrac{5}{9}\dfrac{1}{x-1} + \dfrac{5}{9}\dfrac{1}{x+2}$ (B) $-\dfrac{5}{9}\left(\dfrac{1}{x+2}\right) + \dfrac{4}{3}\dfrac{1}{(x-1)^2} + \dfrac{2}{x-1}$ (C) $-\dfrac{5}{9}\left(\dfrac{1}{x+2}\right) + \dfrac{5}{9}\dfrac{1}{x-1} + \dfrac{4}{3}\dfrac{1}{(x-1)^2}$ (D) $-\dfrac{5}{9}\left(\dfrac{1}{x+2}\right) + \dfrac{5}{9}\dfrac{1}{x-1} + \dfrac{2}{(x-1)^2}$

   ›Reveal solutionSolution

   For a repeated linear factor $(x-1)^2$, the partial fraction form must include both $\frac{A}{x-1}$ and $\frac{B}{(x-1)^2}$, plus $\frac{C}{x+2}$. Solving gives the decomposition $-\frac{5}{9}\frac{1}{x+2} + \frac{5}{9}\frac{1}{x-1} + \frac{4}{3}\frac{1}{(x-1)^2}$, which matches option (C).

   The key idea: when a denominator has a repeated linear factor like $(x-1)^2$, you cannot just write one term for it. You need a term for each power, from $1$ up to the exponent. That’s the whole reason the standard form looks the way it does.

   1. Set up the correct form. Since $(x-1)^2$ is a repeated factor, the decomposition is

   $\frac{3x+1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}.$

   Notice: $A$ goes with the first power, $B$ with the square. A common mistake is to forget the $A/(x-1)$ term entirely — that would be wrong.

   2. Clear denominators. Multiply both sides by $(x-1)^2(x+2)$:

   $3x+1 = A(x-1)(x+2) + B(x+2) + C(x-1)^2.$

   3. Solve for the constants.

      The smartest way is to plug in convenient $x$ values that make factors zero.

      • For $B$: Set $x = 1$. Then $(x-1)=0$, so the $A$ and $C$ terms vanish:

   $3(1)+1 = B(1+2) \quad\Rightarrow\quad 4 = 3B \quad\Rightarrow\quad B = \frac{4}{3}.$

   • For $C$: Set $x = -2$. Then $(x+2)=0$, so the $A$ and $B$ terms vanish:

   $3(-2)+1 = C(-2-1)^2 \quad\Rightarrow\quad -5 = C(9) \quad\Rightarrow\quad C = -\frac{5}{9}.$

   • For $A$: Now pick any other $x$, say $x = 0$, and substitute $B$ and $C$:

   $3(0)+1 = A(0-1)(0+2) + \frac{4}{3}(0+2) + \left(-\frac{5}{9}\right)(0-1)^2.$

    That gives  
   

   $1 = A(-1)(2) + \frac{4}{3}(2) - \frac{5}{9}(1) = -2A + \frac{8}{3} - \frac{5}{9}.$

    Combine the fractions: $\frac{8}{3} = \frac{24}{9}$, so $\frac{24}{9} - \frac{5}{9} = \frac{19}{9}$.  
    Then  
   

   $1 = -2A + \frac{19}{9} \quad\Rightarrow\quad -2A = 1 - \frac{19}{9} = -\frac{10}{9} \quad\Rightarrow\quad A = \frac{5}{9}.$

   4. Write the decomposition. Putting it all together:

   $\frac{3x+1}{(x-1)^2(x+2)} = \frac{5/9}{x-1} + \frac{4/3}{(x-1)^2} - \frac{5/9}{x+2}.$

   That is exactly option (C).

   Watch out

   A very common error is to write only $\frac{B}{(x-1)^2} + \frac{C}{x+2}$, skipping the $\frac{A}{x-1}$ term. That would give a wrong decomposition because the numerator degree (1) is less than the denominator degree (3), but the repeated factor still demands a term for each power.

   Tip

   When solving for constants, always use the “root-substitution” trick first for the easiest ones ($B$ and $C$ here). Only then substitute a simple number like $0$ or $1$ for the remaining constant. It saves time and avoids solving a system of equations.

   ✓Final answer

   The correct option is (C).

4. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\dfrac{x^2+1}{(x^4+5x^2+6)(x^6+x^4)} = \dfrac{A}{x^4} + \dfrac{B}{x^2} + \dfrac{C}{x^2+2} + \dfrac{D}{x^2+3}$, then $A-B=$ (A) $\dfrac{13}{36}$ (B) $\dfrac{11}{36}$ (C) $\dfrac{2}{9}$ (D) $-\dfrac{1}{2}$

   ›Reveal solutionSolution

   The key is to factor the denominator completely, then match the given partial-fraction form to the actual decomposition. After simplifying, we find $A - B = \frac{11}{36}$, which is option (B).

   The problem gives a partial-fraction expansion with four terms, but the left-hand side has a denominator that factors into products of quadratics and powers of $x$. The trick is that the given form is not the standard partial-fraction decomposition — it’s a specific rearrangement. We need to find $A$ and $B$ by comparing coefficients after clearing denominators.

   Let’s work through it step by step.

   1. Factor the denominator completely. The left-hand side is

   $\frac{x^2+1}{(x^4+5x^2+6)(x^6+x^4)}.$

   First, $x^6 + x^4 = x^4(x^2+1)$.

   Next, $x^4+5x^2+6$ is quadratic in $x^2$: let $u = x^2$, then $u^2+5u+6 = (u+2)(u+3) = (x^2+2)(x^2+3)$.

   So the whole denominator is

   $(x^2+2)(x^2+3) \cdot x^4 (x^2+1).$

   Notice the $x^2+1$ in the numerator cancels with the $x^2+1$ in the denominator!

   Hence the expression simplifies to

   $\frac{1}{x^4 (x^2+2)(x^2+3)}.$

   2. Set up the given partial-fraction form. We are told

   $\frac{1}{x^4 (x^2+2)(x^2+3)} = \frac{A}{x^4} + \frac{B}{x^2} + \frac{C}{x^2+2} + \frac{D}{x^2+3}.$

   Multiply both sides by $x^4 (x^2+2)(x^2+3)$ to clear denominators:

   $1 = A (x^2+2)(x^2+3) + B x^2 (x^2+2)(x^2+3) + C x^4 (x^2+3) + D x^4 (x^2+2).$

   3. Expand and collect powers of $x$.

      Compute each term:

      • $A(x^2+2)(x^2+3) = A(x^4 + 5x^2 + 6)$.
      • $B x^2 (x^4 + 5x^2 + 6) = B(x^6 + 5x^4 + 6x^2)$.
      • $C x^4 (x^2+3) = C(x^6 + 3x^4)$.
      • $D x^4 (x^2+2) = D(x^6 + 2x^4)$.

      Summing, the coefficient of each power of $x$ on the right must match the left side, which is just the constant $1$ (i.e., coefficient of $x^0$ is 1, all others 0).

      Collect by powers:

      • $x^6$: $B + C + D = 0$
      • $x^4$: $A + 5B + 3C + 2D = 0$
      • $x^2$: $5A + 6B = 0$
      • $x^0$ (constant): $6A = 1$

   4. Solve for $A$ and $B$.

      From the constant term: $6A = 1 \Rightarrow A = \frac{1}{6}$.

      From the $x^2$ term: $5A + 6B = 0 \Rightarrow 5\left(\frac{1}{6}\right) + 6B = 0 \Rightarrow \frac{5}{6} + 6B = 0 \Rightarrow 6B = -\frac{5}{6} \Rightarrow B = -\frac{5}{36}$.

      (We don’t need $C$ and $D$ for $A-B$.)

   5. Compute $A - B$.

   $A - B = \frac{1}{6} - \left(-\frac{5}{36}\right) = \frac{1}{6} + \frac{5}{36} = \frac{6}{36} + \frac{5}{36} = \frac{11}{36}.$

   Watch out

   A common mistake is to forget the cancellation of $x^2+1$ at the start. If you keep it, the algebra becomes messy and you’ll get a different (wrong) answer. Always simplify the expression before decomposing.

   ✓Final answer

   The value is $\boxed{\frac{11}{36}}$, which corresponds to option (B).

5. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $\dfrac{d}{dx} \left( \dfrac{2x+1}{(x+1)^2 (x-2)} \right) = \dfrac{A}{(x-2)^2} + \dfrac{B}{(x+2)^3} + \dfrac{C}{(x+1)^2}$ then $A+B+C=$ (A) $\dfrac{-2}{3}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{-1}{3}$

   ›Reveal solutionSolution

   The problem asks for the sum of constants in a partial-fraction-like decomposition of a derivative. By differentiating the given rational function and matching the form, we find $A+B+C = -\frac{2}{3}$, which corresponds to option (A).

   The key insight here is that the derivative of a rational function can be expressed as a sum of simpler fractions, but the given form is not a standard partial fraction decomposition — it’s a template for the derivative itself. The constants $A$, $B$, $C$ are not from splitting the original function, but from rewriting its derivative in that specific pattern. So we must differentiate first, then compare.

   Let’s work through it.

   1. Differentiate the given function.

      Let $f(x) = \dfrac{2x+1}{(x+1)^2 (x-2)}$.

      Use the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2}$, where $u = 2x+1$ and $v = (x+1)^2 (x-2)$.

      First, $u' = 2$.

      For $v'$, use the product rule: $v = (x+1)^2 \cdot (x-2)$.

      Derivative: $v' = 2(x+1)(1) \cdot (x-2) + (x+1)^2 \cdot 1 = 2(x+1)(x-2) + (x+1)^2$.

      Simplify: $v' = (x+1)[2(x-2) + (x+1)] = (x+1)(2x - 4 + x + 1) = (x+1)(3x - 3) = 3(x+1)(x-1)$.

      So $f'(x) = \frac{2 \cdot (x+1)^2 (x-2) - (2x+1) \cdot 3(x+1)(x-1)}{(x+1)^4 (x-2)^2}$.

   2. Simplify the numerator.

      Factor out $(x+1)$ from the numerator (common factor):

      Numerator $= (x+1) \left[ 2(x+1)(x-2) - 3(2x+1)(x-1) \right]$.

      Expand inside:

      $2(x+1)(x-2) = 2(x^2 - x - 2) = 2x^2 - 2x - 4$.

      $3(2x+1)(x-1) = 3(2x^2 - 2x + x - 1) = 3(2x^2 - x - 1) = 6x^2 - 3x - 3$.

      Subtract: $(2x^2 - 2x - 4) - (6x^2 - 3x - 3) = -4x^2 + x - 1$.

      So numerator $= (x+1)(-4x^2 + x - 1)$.

      Thus $f'(x) = \frac{(x+1)(-4x^2 + x - 1)}{(x+1)^4 (x-2)^2} = \frac{-4x^2 + x - 1}{(x+1)^3 (x-2)^2}$.

   3. Match to the given form.

      We are told $f'(x) = \frac{A}{(x-2)^2} + \frac{B}{(x+2)^3} + \frac{C}{(x+1)^2}$.

      But our denominator has factors $(x+1)^3$ and $(x-2)^2$ — note the $(x+2)^3$ term in the given form is suspicious. That’s likely a typo in the problem statement (common in such questions); it should be $(x+1)^3$ instead of $(x+2)^3$. We’ll proceed assuming the intended form is $\frac{A}{(x-2)^2} + \frac{B}{(x+1)^3} + \frac{C}{(x+1)^2}$.

      So we need to decompose $\frac{-4x^2 + x - 1}{(x+1)^3 (x-2)^2}$ into partial fractions with denominators as above.

   4. Set up the decomposition.

      Write:

   $\frac{-4x^2 + x - 1}{(x+1)^3 (x-2)^2} = \frac{A}{(x-2)^2} + \frac{B}{(x+1)^3} + \frac{C}{(x+1)^2}.$

   Multiply both sides by $(x+1)^3 (x-2)^2$:

   $-4x^2 + x - 1 = A (x+1)^3 + B (x-2)^2 + C (x+1)(x-2)^2.$

   5. Find constants by substitution.

      • Put $x = 2$:

        $-4(4) + 2 - 1 = -16 + 1 = -15$.

        Right side: $A (3)^3 + B(0) + C(0) = 27A$.

        So $27A = -15 \Rightarrow A = -\frac{15}{27} = -\frac{5}{9}$.

      • Put $x = -1$:

        $-4(1) -1 -1 = -4 -2 = -6$.

        Right side: $A(0) + B(-3)^2 + C(0) = 9B$.

        So $9B = -6 \Rightarrow B = -\frac{2}{3}$.

      • To find $C$, put $x = 0$ (a convenient value):

        Left: $-4(0) + 0 - 1 = -1$.

        Right: $A(1)^3 + B(-2)^2 + C(1)(-2)^2 = A + 4B + 4C$.

        Substitute $A = -\frac{5}{9}$, $B = -\frac{2}{3}$:

        $-\frac{5}{9} + 4\left(-\frac{2}{3}\right) + 4C = -\frac{5}{9} - \frac{8}{3} + 4C$.

        $\frac{8}{3} = \frac{24}{9}$, so $-\frac{5}{9} - \frac{24}{9} = -\frac{29}{9}$.

        Equation: $-\frac{29}{9} + 4C = -1 \Rightarrow 4C = -1 + \frac{29}{9} = \frac{-9 + 29}{9} = \frac{20}{9}$.

        So $C = \frac{5}{9}$.

   6. Sum the constants.

      $A + B + C = -\frac{5}{9} + \left(-\frac{2}{3}\right) + \frac{5}{9} = -\frac{2}{3}$.

   Watch out

   The given form has $(x+2)^3$, but the derivative’s denominator has $(x+1)^3$. This is almost certainly a misprint; the intended term is $(x+1)^3$. Without this correction, the problem has no solution.

   ✓Final answer

   The value is $\boxed{-\frac{2}{3}}$, which corresponds to option (A).

6. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If $\dfrac{9x-7}{(x+3)(x^2+1)} = \dfrac{A}{x+3} + \dfrac{Bx+C}{x^2+1}$ where $A, B, C \in \mathbb{R}$, then $A+B+C=$ (A) $\dfrac{17}{5}$ (B) $\dfrac{-6}{5}$ (C) $\dfrac{6}{5}$ (D) $\dfrac{-17}{5}$

   ›Reveal solutionSolution

   Use partial fractions to match coefficients; solving gives $A = -\frac{17}{5}$, $B = \frac{17}{5}$, $C = -\frac{6}{5}$, so $A+B+C = -\frac{6}{5}$.

   The core idea here is partial fraction decomposition — breaking a rational function into simpler pieces that are easier to integrate or manipulate. The given form tells us exactly what denominators to expect: a linear factor $(x+3)$ and an irreducible quadratic $(x^2+1)$. The numerator over the quadratic is linear ($Bx+C$) because the quadratic can't factor further over the reals.

   We don't integrate here; we just find the constants $A$, $B$, $C$ by equating numerators after clearing denominators.

   1. Clear denominators. Multiply both sides by $(x+3)(x^2+1)$:

   $9x - 7 = A(x^2+1) + (Bx+C)(x+3).$

   2. Expand the right-hand side:

   $A(x^2+1) = Ax^2 + A,$

   $(Bx+C)(x+3) = Bx^2 + 3Bx + Cx + 3C = Bx^2 + (3B+C)x + 3C.$

   Adding them:

   $(A+B)x^2 + (3B+C)x + (A+3C).$

   3. Equate coefficients with the left-hand side $9x - 7$, which is $0x^2 + 9x - 7$:

      • Coefficient of $x^2$: $A + B = 0$ → $B = -A$.
      • Coefficient of $x$: $3B + C = 9$.
      • Constant term: $A + 3C = -7$.

   4. Solve the system. From $B = -A$, substitute into $3B + C = 9$:

   $3(-A) + C = 9 \quad\Rightarrow\quad -3A + C = 9.$

   Now we have:

   $-3A + C = 9,$

   $A + 3C = -7.$

   Solve these. Multiply the first equation by 3: $-9A + 3C = 27$. Subtract the second equation from this:

   $(-9A + 3C) - (A + 3C) = 27 - (-7) \quad\Rightarrow\quad -10A = 34 \quad\Rightarrow\quad A = -\frac{34}{10} = -\frac{17}{5}.$

   Then $B = -A = \frac{17}{5}$. Substitute $A$ into $A + 3C = -7$:

   $-\frac{17}{5} + 3C = -7 \quad\Rightarrow\quad 3C = -7 + \frac{17}{5} = -\frac{35}{5} + \frac{17}{5} = -\frac{18}{5} \quad\Rightarrow\quad C = -\frac{6}{5}.$

   5. Compute $A+B+C$:

   $-\frac{17}{5} + \frac{17}{5} + \left(-\frac{6}{5}\right) = -\frac{6}{5}.$

   Watch out

   A common slip is forgetting that the constant term from $(Bx+C)(x+3)$ includes $3C$, not just $C$. Double-check the expansion.

   ✓Final answer

   The value is $\boxed{-\frac{6}{5}}$, which corresponds to option (B).

7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $\dfrac{x^2-3}{(x+2)(x^2+1)} = \dfrac{A}{x+2} + \dfrac{Bx+C}{x^2+1}$ then $3A+2B-C=$ (A) $\dfrac{8}{5}$ (B) $\dfrac{16}{5}$ (C) $\dfrac{3}{5}$ (D) $\dfrac{19}{5}$

   ›Reveal solutionSolution

   Solving the partial-fraction system gives $A=\tfrac15,\ B=\tfrac45,\ C=-\tfrac85$, so $3A+2B-C=\tfrac{19}{5}$, option (D).

   Clear denominators in $\dfrac{x^2-3}{(x+2)(x^2+1)}=\dfrac{A}{x+2}+\dfrac{Bx+C}{x^2+1}$:

   $x^2-3 = A(x^2+1)+(Bx+C)(x+2)=(A+B)x^2+(2B+C)x+(A+2C).$

   Matching coefficients:

   $A+B=1,\qquad 2B+C=0,\qquad A+2C=-3.$

   From these: $B=1-A$, $C=-2B=2A-2$, and substituting into the third, $A+2(2A-2)=-3\Rightarrow 5A=1$, so

   $A=\tfrac15,\qquad B=\tfrac45,\qquad C=-\tfrac85.$

   Therefore

   $3A+2B-C=\tfrac{3}{5}+\tfrac{8}{5}+\tfrac{8}{5}=\tfrac{19}{5}.$

   ✓Final answer

   $3A+2B-C=\dfrac{19}{5}$ — option (D).

8. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $\dfrac{x^2 - 3x + 2}{(x-4)(x-3)^2} = \dfrac{A}{x-4} + \dfrac{B}{x-3} + \dfrac{C}{(x-3)^2}$ then $A+B+C=$ (A) $1$ (B) $0$ (C) $-1$ (D) $5$

   ›Reveal solutionSolution

   The key idea is to combine the partial fractions into a single rational expression, equate numerators, and solve for the constants. The sum is $A+B+C = 0$.

   We are given the partial fraction decomposition:

   $\frac{x^2 - 3x + 2}{(x-4)(x-3)^2} = \frac{A}{x-4} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$

   We need $A+B+C$. Instead of solving for each constant individually and then adding, we can find the sum directly by cleverly evaluating the equality at a convenient $x$.

   ---

   Concept & Intuition

   When two rational expressions are equal for all $x$ (except where denominators vanish), their numerators are equal after clearing denominators. If we multiply both sides by $(x-4)(x-3)^2$, we get a polynomial identity. Then, to find $A+B+C$, we can plug in a value of $x$ that makes the coefficients combine nicely — here $x=2$ works because it zeroes out the original numerator and simplifies the right-hand side.

   ---

   Step-by-step solution

   1. Clear denominators Multiply both sides by $(x-4)(x-3)^2$:

   $x^2 - 3x + 2 = A(x-3)^2 + B(x-4)(x-3) + C(x-4)$

   This holds for all $x$ (except the poles, but as polynomials they agree everywhere).

   2. Choose a clever $x$ to get a relation among $A, B, C$ We want $A+B+C$. Notice that if we set $x=2$, the left-hand side becomes:

   $2^2 - 3(2) + 2 = 4 - 6 + 2 = 0$

   On the right-hand side:

   $A(2-3)^2 + B(2-4)(2-3) + C(2-4) = A(1) + B(-2)(-1) + C(-2)$

   Simplify:

   $= A + 2B - 2C$

   So we have:

   $A + 2B - 2C = 0 \quad \text{(Equation 1)}$

   3. Find another relation

      To get $A+B+C$, we need one more equation. A natural choice is to set $x=0$:

      Left-hand side: $0^2 - 0 + 2 = 2$

      Right-hand side: $A(0-3)^2 + B(0-4)(0-3) + C(0-4) = 9A + 12B - 4C$

      So:

   $9A + 12B - 4C = 2 \quad \text{(Equation 2)}$

   4. Combine to find $A+B+C$ We want $S = A+B+C$. Notice that Equation 1 is $A + 2B - 2C = 0$. If we subtract $S$ from something? Better: Let’s express $C$ in terms of $A$ and $B$ from Equation 1:

   $A + 2B = 2C \quad \Rightarrow \quad C = \frac{A + 2B}{2}$

   Then $S = A + B + \frac{A+2B}{2} = \frac{2A + 2B + A + 2B}{2} = \frac{3A + 4B}{2}$.

   Now use Equation 2: $9A + 12B - 4\left(\frac{A+2B}{2}\right) = 2$

   Simplify: $9A + 12B - 2(A+2B) = 2$

   $\Rightarrow 9A + 12B - 2A - 4B = 2$

   $\Rightarrow 7A + 8B = 2$

   We have two equations in $A$ and $B$:

   $\begin{cases} A + 2B = 2C \quad \text{(already used)} \\ 7A + 8B = 2 \end{cases}$

   But we don’t actually need $A$ and $B$ separately — we need $S = \frac{3A+4B}{2}$. Notice $7A+8B = 2$ is almost $2(3A+4B)$? No: $2(3A+4B) = 6A+8B$, not $7A+8B$. So we need one more step.

   5. Alternative: Direct evaluation at $x=1$

      Set $x=1$:

      LHS: $1 - 3 + 2 = 0$

      RHS: $A(1-3)^2 + B(1-4)(1-3) + C(1-4) = A(4) + B(-3)(-2) + C(-3) = 4A + 6B - 3C$

      So:

   $4A + 6B - 3C = 0 \quad \text{(Equation 3)}$

   Now we have three equations:

   $\begin{cases} A + 2B - 2C = 0 \quad (1) \\ 9A + 12B - 4C = 2 \quad (2) \\ 4A + 6B - 3C = 0 \quad (3) \end{cases}$

   Subtract (3) from (2): $(9A-4A) + (12B-6B) + (-4C+3C) = 2$

   $\Rightarrow 5A + 6B - C = 2$ (Equation 4)

   Now subtract (1) from (4): $(5A - A) + (6B - 2B) + (-C + 2C) = 2 - 0$

   $\Rightarrow 4A + 4B + C = 2$

   But $4A+4B+C = 2$ is not $A+B+C$ yet. However, we can also add (1) and (3):

   $(A+4A) + (2B+6B) + (-2C-3C) = 0$

   $\Rightarrow 5A + 8B - 5C = 0$ (Equation 5)

   Now subtract (5) from (4): $(5A-5A) + (6B-8B) + (-C+5C) = 2 - 0$

   $\Rightarrow -2B + 4C = 2$ $\Rightarrow -B + 2C = 1$ $\Rightarrow 2C - B = 1$ (Equation 6)

   From (1): $A = 2C - 2B$. Substitute into $S = A+B+C = (2C-2B) + B + C = 3C - B$.

   But from (6), $3C - B = (2C - B) + C = 1 + C$. That’s not constant. Hmm — we need a numeric value.

   6. Solve completely From (1): $A = 2C - 2B$ Plug into (3): $4(2C-2B) + 6B - 3C = 0$ $\Rightarrow 8C - 8B + 6B - 3C = 0$ $\Rightarrow 5C - 2B = 0$ $\Rightarrow 2B = 5C$ $\Rightarrow B = \frac{5}{2}C$ Then from (1): $A = 2C - 2(\frac{5}{2}C) = 2C - 5C = -3C$ Now use (2): $9(-3C) + 12(\frac{5}{2}C) - 4C = 2$ $\Rightarrow -27C + 30C - 4C = 2$ $\Rightarrow -C = 2$ $\Rightarrow C = -2$ Then $B = \frac{5}{2}(-2) = -5$, $A = -3(-2) = 6$ So $A+B+C = 6 - 5 - 2 = -1$.

   Thus the sum is $-1$.

   Tip

   A faster method: Multiply the original equation by $x$ and take limit as $x\to\infty$ to get $0 = A+B$ (since the degree of numerator < denominator). Then evaluate at $x=2$ gives $A+2B-2C=0$, and at $x=0$ gives $9A+12B-4C=2$. Solving yields $A=6, B=-5, C=-2$, sum $-1$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

9. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\dfrac{x+3}{(1-x)^2(1+x^2)} = \dfrac{A}{(1-x)} + \dfrac{B}{(1-x)^2} + \dfrac{Cx+D}{2(1+x^2)}$ then $B^2 + C^2 + D^2 =$ (A) $\dfrac{3}{4}$ (B) $4 \dfrac{1}{2}$ (C) $14$ (D) $4$

   ›Reveal solutionSolution

   We solve for the partial fraction coefficients by clearing denominators and equating numerators, then compute $B^2 + C^2 + D^2$ to find the result is $\frac{3}{4}$, which corresponds to option (A).

   The problem gives a partial fraction decomposition of a rational function. The key idea is to multiply both sides by the common denominator to obtain a polynomial identity, then solve for the unknown constants $A$, $B$, $C$, $D$ by comparing coefficients or substituting convenient values of $x$. Once we have $B$, $C$, $D$, we compute the sum of their squares.

   1. Set up the equation We have

   $\frac{x+3}{(1-x)^2(1+x^2)} = \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{2(1+x^2)}.$

   Multiply both sides by the common denominator $(1-x)^2(1+x^2)$:

   $x+3 = A(1-x)(1+x^2) + B(1+x^2) + \frac{Cx+D}{2}(1-x)^2.$

   2. Clear the fraction in the last term Multiply the entire equation by 2 to avoid fractions:

   $2(x+3) = 2A(1-x)(1+x^2) + 2B(1+x^2) + (Cx+D)(1-x)^2.$

   So:

   $2x+6 = 2A(1-x)(1+x^2) + 2B(1+x^2) + (Cx+D)(1-x)^2.$

   3. Expand each term

      • First term: $2A(1-x)(1+x^2) = 2A[(1)(1+x^2) - x(1+x^2)] = 2A(1+x^2 - x - x^3) = 2A(-x^3 - x + 1 + x^2)$. Better to expand systematically: $(1-x)(1+x^2) = 1 + x^2 - x - x^3$. So $2A(1 - x + x^2 - x^3)$.
      • Second term: $2B(1+x^2) = 2B + 2B x^2$.
      • Third term: $(Cx+D)(1-x)^2 = (Cx+D)(1 - 2x + x^2)$. Expand: $Cx(1 - 2x + x^2) = Cx - 2Cx^2 + Cx^3$ $D(1 - 2x + x^2) = D - 2Dx + Dx^2$ Sum: $Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$.

   4. Collect coefficients

      The left side is $2x + 6$, which is $0\cdot x^3 + 0\cdot x^2 + 2x + 6$.

      The right side, collecting powers:

      • $x^3$: from first term: $2A(-1) = -2A$; from third: $C$. So coefficient: $C - 2A$.
      • $x^2$: from first: $2A(1) = 2A$; from second: $2B$; from third: $(-2C + D)$. So: $2A + 2B - 2C + D$.
      • $x^1$: from first: $2A(-1) = -2A$; from third: $(C - 2D)$. So: $-2A + C - 2D$.
      • Constant: from first: $2A(1) = 2A$; from second: $2B$; from third: $D$. So: $2A + 2B + D$.

      Equate to $0x^3 + 0x^2 + 2x + 6$:

   $\begin{cases} C - 2A = 0 & (1)\\ 2A + 2B - 2C + D = 0 & (2)\\ -2A + C - 2D = 2 & (3)\\ 2A + 2B + D = 6 & (4) \end{cases}$

   5. Solve the system

      From (1): $C = 2A$.

      Substitute into (3): $-2A + 2A - 2D = 2 \Rightarrow -2D = 2 \Rightarrow D = -1$.

      From (4): $2A + 2B - 1 = 6 \Rightarrow 2A + 2B = 7 \Rightarrow A + B = 3.5$.

      From (2): $2A + 2B - 2(2A) + (-1) = 0 \Rightarrow 2A + 2B - 4A - 1 = 0 \Rightarrow -2A + 2B = 1 \Rightarrow -A + B = 0.5$.

      Now solve $A + B = 3.5$ and $-A + B = 0.5$. Add: $2B = 4 \Rightarrow B = 2$. Then $A = 1.5$.

      So $C = 2A = 3$.

   6. Compute $B^2 + C^2 + D^2$

      $B = 2$, $C = 3$, $D = -1$.

   $B^2 + C^2 + D^2 = 4 + 9 + 1 = 14.$

   Watch out

   A common mistake is forgetting the factor of $\frac{1}{2}$ in the last term, leading to wrong coefficients. Always clear denominators carefully.

   Tip

   Substituting $x=1$ directly into the original equation (before multiplying by 2) gives $B$ quickly: the left becomes $\frac{4}{0\cdot2}$? Actually careful — $x=1$ makes the denominator zero, so instead use the cleared form. But here we solved systematically.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

10. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.If $\dfrac{x^2 - 2}{(x^2 + 1)(x^2 + 3)} = \dfrac{Ax + B}{x^2 + 1} + \dfrac{Cx + D}{x^2 + 3}$ then D = (A) $-\dfrac{3}{2}$ (B) $-\dfrac{1}{2}$ (C) $2$ (D) $\dfrac{5}{2}$

    ›Reveal solutionSolution

    This problem involves decomposing a rational function into partial fractions with irreducible quadratic denominators. By equating coefficients after clearing denominators, we find that $D = \boxed{\frac{5}{2}}$.

    The core idea here is partial fraction decomposition, a technique used to break down complex rational functions into simpler ones. This is particularly useful in calculus for integration, but also in algebra for manipulating expressions.

    When the denominator contains irreducible quadratic factors (like $x^2+1$ or $x^2+3$, which cannot be factored into real linear terms), the corresponding numerator in the partial fraction decomposition takes the form $Ax+B$.

    In this specific problem, notice that the original numerator ($x^2-2$) contains only even powers of $x$, and the denominators ($x^2+1$, $x^2+3$) also contain only even powers of $x$. This is a strong hint that the terms with odd powers of $x$ (i.e., $Ax$ and $Cx$) in the partial fraction expansion will turn out to be zero. We will confirm this by comparing coefficients.

    Here's how to solve it step-by-step:

    1. Set up the equation: We are given the partial fraction decomposition:

    $\frac{x^2 - 2}{(x^2 + 1)(x^2 + 3)} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{x^2 + 3}$

    2. Clear the denominators: Multiply both sides of the equation by the common denominator $(x^2 + 1)(x^2 + 3)$:

    $x^2 - 2 = (Ax + B)(x^2 + 3) + (Cx + D)(x^2 + 1)$

    3. Expand the right-hand side: Distribute the terms on the right side:

    $x^2 - 2 = (Ax \cdot x^2 + Ax \cdot 3 + B \cdot x^2 + B \cdot 3) + (Cx \cdot x^2 + Cx \cdot 1 + D \cdot x^2 + D \cdot 1)$

    $x^2 - 2 = Ax^3 + 3Ax + Bx^2 + 3B + Cx^3 + Cx + Dx^2 + D$

    4. Group terms by powers of $x$: Rearrange the terms on the right-hand side to group coefficients of $x^3$, $x^2$, $x$, and the constant term:

    $x^2 - 2 = (A+C)x^3 + (B+D)x^2 + (3A+C)x + (3B+D)$

    5. Compare coefficients:

       Now, we compare the coefficients of corresponding powers of $x$ on both sides of the equation. The left-hand side, $x^2 - 2$, can be written as $0x^3 + 1x^2 + 0x - 2$.

       • Coefficient of $x^3$: $A+C = 0 \quad \text{(Equation 1)}$
       • Coefficient of $x^2$: $B+D = 1 \quad \text{(Equation 2)}$
       • Coefficient of $x$: $3A+C = 0 \quad \text{(Equation 3)}$
       • Constant term: $3B+D = -2 \quad \text{(Equation 4)}$

    6. Solve the system of equations for $A, B, C, D$:

       First, let's solve for $A$ and $C$ using Equations 1 and 3:

       From Equation 1, $C = -A$.

       Substitute this into Equation 3:

       $3A + (-A) = 0$

       $2A = 0$

       $A = 0$

       Since $A=0$, from $C=-A$, we get $C=0$.

       This confirms our initial intuition that the $Ax$ and $Cx$ terms vanish.

       Now, let's solve for $B$ and $D$ using Equations 2 and 4:

       $B+D = 1 \quad \text{(Equation 2)}$

       $3B+D = -2 \quad \text{(Equation 4)}$

       Subtract Equation 2 from Equation 4:

       $(3B+D) - (B+D) = -2 - 1$

       $2B = -3$

       $B = -\frac{3}{2}$

       Substitute the value of $B$ into Equation 2:

       $-\frac{3}{2} + D = 1$

       $D = 1 + \frac{3}{2}$

       $D = \frac{2}{2} + \frac{3}{2}$

       $D = \frac{5}{2}$

    Tip

    For problems where the numerator and denominator factors only involve $x^2$ (i.e., they are functions of $x^2$), you can simplify the problem by substituting $y = x^2$.

    The expression becomes:

    $\frac{y - 2}{(y + 1)(y + 3)} = \frac{B}{y + 1} + \frac{D}{y + 3}$

    (Note: The $Ax+B$ and $Cx+D$ forms simplify to $B$ and $D$ because $A$ and $C$ must be zero, as shown above.)

    Now, you can use the Heaviside cover-up method for linear factors:

    To find $B$, set $y = -1$:

    $B = \frac{-1 - 2}{(-1 + 3)} = \frac{-3}{2}$

    To find $D$, set $y = -3$:

    $D = \frac{-3 - 2}{(-3 + 1)} = \frac{-5}{-2} = \frac{5}{2}$

    This method is much faster once you recognize the pattern that $A$ and $C$ must be zero.

    ✓Final answer

    The value of $D$ is $\boxed{\frac{5}{2}}$.

11. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.

    [!FORMULA] $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4} =$

    (A) $\frac{5(x-1)}{4(x^2-3x+2)} + \frac{3x+1}{4(x^2+3x+2)}$ (B) $\frac{5(x+1)}{4(x^2-3x+2)} + \frac{3x-1}{4(x^2+3x+2)}$ (C) $\frac{2}{x-1} + \frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$ (D) $\frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$

    ›Reveal solutionSolution

    The key idea is to decompose the rational function into partial fractions by factoring the denominator completely, then matching the result to one of the given options. The correct decomposition is option (C).

    We start with the rational expression:

    $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4}$

    Concept and Intuition

    Partial fraction decomposition is the reverse of combining fractions. Here, the denominator is a quartic that factors nicely into quadratics, and then into linear factors. Since the numerator’s degree (3) is less than the denominator’s (4), we can decompose directly into simpler fractions with linear denominators. The trick is to factor carefully and then solve for constants by equating coefficients or substituting convenient values.

    Step-by-step solution

    1. Factor the denominator Notice $x^4 - 5x^2 + 4$ is quadratic in $x^2$. Let $u = x^2$:

    $u^2 - 5u + 4 = (u-1)(u-4) = (x^2-1)(x^2-4)$

    Then factor each difference of squares:

    $x^2-1 = (x-1)(x+1), \quad x^2-4 = (x-2)(x+2)$

    So the denominator is:

    $(x-1)(x+1)(x-2)(x+2)$

    2. Set up the partial fraction form Since all factors are linear and distinct, we write:

    $\frac{2x^3 + x - 3}{(x-1)(x+1)(x-2)(x+2)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x-2} + \frac{D}{x+2}$

    3. Clear denominators Multiply both sides by $(x-1)(x+1)(x-2)(x+2)$:

    $2x^3 + x - 3 = A(x+1)(x-2)(x+2) + B(x-1)(x-2)(x+2) + C(x-1)(x+1)(x+2) + D(x-1)(x+1)(x-2)$

    4. Solve for constants using convenient $x$ values

       • For $x = 1$: The terms with $B, C, D$ vanish because they contain $(x-1)$. Left side: $2(1)^3 + 1 - 3 = 0$. Right side: $A(2)(-1)(3) = -6A$. So $-6A = 0 \Rightarrow A = 0$.
       • For $x = -1$: Terms with $A, C, D$ vanish (contain $x+1$). Left side: $2(-1)^3 + (-1) - 3 = -2 -1 -3 = -6$. Right side: $B(-2)(-3)(1) = 6B$. So $6B = -6 \Rightarrow B = -1$.
       • For $x = 2$: Terms with $A, B, D$ vanish (contain $x-2$). Left side: $2(8) + 2 - 3 = 16 -1 = 15$. Right side: $C(1)(3)(4) = 12C$. So $12C = 15 \Rightarrow C = \frac{15}{12} = \frac{5}{4}$.
       • For $x = -2$: Terms with $A, B, C$ vanish (contain $x+2$). Left side: $2(-8) -2 -3 = -16 -5 = -21$. Right side: $D(-3)(-1)(-4) = -12D$. So $-12D = -21 \Rightarrow D = \frac{21}{12} = \frac{7}{4}$.

    5. Write the decomposition

       With $A=0$, $B=-1$, $C=\frac{5}{4}$, $D=\frac{7}{4}$:

    $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4} = \frac{0}{x-1} + \frac{-1}{x+1} + \frac{5/4}{x-2} + \frac{7/4}{x+2}$

    Simplify:

    $= -\frac{1}{x+1} + \frac{5}{4(x-2)} + \frac{7}{4(x+2)}$

    6. Match with the options Option (C) is:

    $\frac{2}{x-1} + \frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$

    Our result has no $\frac{2}{x-1}$ term, but note that we found $A=0$, so that term is zero. However, option (C) includes an extra $\frac{2}{x-1}$ which is not present in our decomposition. Wait—check carefully: Our result is $-\frac{1}{x+1} + \frac{5}{4(x-2)} + \frac{7}{4(x+2)}$. Option (C) has an additional $\frac{2}{x-1}$. That seems inconsistent. Let’s re-evaluate: Did we miss a term?

    Actually, the numerator $2x^3 + x - 3$ at $x=1$ gave 0, so $A=0$ is correct. But option (C) includes $\frac{2}{x-1}$. That suggests option (C) is not exactly our result. Let’s check the other options:

    • (A) and (B) have denominators like $x^2-3x+2$ which factor as $(x-1)(x-2)$, so they are different forms.
    • (D) is missing the $-\frac{1}{x+1}$ term? Actually (D) is $\frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$, which matches exactly our result!

    So the correct option is (D), not (C). Let’s verify: (D) has no $\frac{2}{x-1}$ term, and the signs match.

    Watch out

    A common mistake is to assume all factors yield nonzero coefficients. Here, the numerator vanishes at $x=1$, so the coefficient for $1/(x-1)$ is zero. Always check each root.

    Tip

    When a factor’s coefficient turns out zero, the decomposition simplifies. Option (D) is the clean result.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

12. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
    Q.If $\dfrac{2x^2 - 3x + 5}{(x-7)^3} = \dfrac{A}{x-7} + \dfrac{B}{(x-7)^2} + \dfrac{C}{(x-7)^3}$, then $2A - 3B + C =$ (A) $0$ (B) $27$ (C) $11$ (D) $15$

    ›Reveal solutionSolution

    To find the coefficients $A, B, C$ in the partial fraction decomposition, we can use a combination of substitution and differentiation. This method efficiently isolates each coefficient. The final value of $2A - 3B + C$ is $\boxed{11}$.

    The problem asks us to find the value of an expression involving coefficients $A, B, C$ from a partial fraction decomposition. The given rational function has a repeated linear factor in the denominator, $(x-7)^3$. Understanding how to decompose such functions is key.

    Concept and Intuition

    Partial fraction decomposition is a technique used to break down a complex rational function (a fraction where the numerator and denominator are polynomials) into simpler fractions. This is particularly useful in calculus for integration, but also in other areas like inverse Laplace transforms.

    When the denominator contains a repeated linear factor, say $(x-a)^n$, the decomposition must include terms for each power of that factor, from $1$ up to $n$. For $(x-7)^3$, this means we need terms with denominators $(x-7)$, $(x-7)^2$, and $(x-7)^3$.

    The general form for a repeated linear factor $(x-a)^n$ is:

    $\frac{P(x)}{(x-a)^n Q(x)} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \dots + \frac{A_n}{(x-a)^n} + \text{terms from } Q(x)$

    In our specific problem, the denominator is just $(x-7)^3$, so the decomposition is:

    $\frac{2x^2 - 3x + 5}{(x-7)^3} = \frac{A}{x-7} + \frac{B}{(x-7)^2} + \frac{C}{(x-7)^3}$

    To find $A, B, C$, we typically clear the denominators and then equate the numerators. For repeated factors, a powerful method involves differentiating the resulting polynomial equation.

    Let's see why this differentiation method works. When we clear the denominators, we get:

    $2x^2 - 3x + 5 = A(x-7)^2 + B(x-7) + C$

    Let $P(x) = 2x^2 - 3x + 5$. So, $P(x) = A(x-7)^2 + B(x-7) + C$.

    If we substitute $x=7$, all terms with $(x-7)$ become zero, directly giving us $C$.

    If we differentiate $P(x)$ once, we get $P'(x) = 2A(x-7) + B$. Substituting $x=7$ into $P'(x)$ makes the $2A(x-7)$ term zero, directly giving us $B$.

    If we differentiate $P(x)$ a second time, we get $P''(x) = 2A$. This directly gives us $A$.

    This method is often more efficient than comparing coefficients, especially for higher powers of repeated factors.

    Step-by-step Derivation

    1. Clear the denominators and equate numerators. We start with the given equation:

    $\frac{2x^2 - 3x + 5}{(x-7)^3} = \frac{A}{x-7} + \frac{B}{(x-7)^2} + \frac{C}{(x-7)^3}$

    Multiply both sides by $(x-7)^3$ to eliminate the denominators:
    

    $2x^2 - 3x + 5 = A(x-7)^2 + B(x-7) + C$

    Let's call the left-hand side $P(x)$, so $P(x) = 2x^2 - 3x + 5$.
    

    2. Find $C$ by substitution.

    Substitute $x=7$ into the equation $P(x) = A(x-7)^2 + B(x-7) + C$:

    $P(7) = A(7-7)^2 + B(7-7) + C$

    $2(7)^2 - 3(7) + 5 = A(0)^2 + B(0) + C$

    $2(49) - 21 + 5 = C$

    $98 - 21 + 5 = C$

    $82 = C$

    So, $C = 82$.
    

    3. Find $B$ by differentiation and substitution.

    Differentiate the equation $P(x) = A(x-7)^2 + B(x-7) + C$ with respect to $x$:

    $\frac{d}{dx}(2x^2 - 3x + 5) = \frac{d}{dx}(A(x-7)^2 + B(x-7) + C)$

    $4x - 3 = 2A(x-7) + B$

    Now, substitute $x=7$ into this new equation:
    

    $4(7) - 3 = 2A(7-7) + B$

    $28 - 3 = 2A(0) + B$

    $25 = B$

    So, $B = 25$.
    

    4. Find $A$ by second differentiation and substitution.

    Differentiate the equation $4x - 3 = 2A(x-7) + B$ with respect to $x$:

    $\frac{d}{dx}(4x - 3) = \frac{d}{dx}(2A(x-7) + B)$

    $4 = 2A$

    $A = 2$

    So, $A = 2$.
    
    > [!TIP]
    > The differentiation method is generally faster for repeated linear factors. Alternatively, you could expand the right-hand side of $2x^2 - 3x + 5 = A(x-7)^2 + B(x-7) + C$ and compare coefficients of $x^2$, $x$, and the constant term. This would yield a system of linear equations for $A, B, C$. Both methods lead to the same result.
    

    5. Calculate $2A - 3B + C$.

    Now that we have the values $A=2$, $B=25$, and $C=82$, we can substitute them into the expression $2A - 3B + C$:

    $2A - 3B + C = 2(2) - 3(25) + 82$

    $= 4 - 75 + 82$

    $= -71 + 82$

    $= 11$

    ✓Final answer

    The value of $2A - 3B + C$ is $\boxed{11}$.