Q.Find $\int \dfrac{dx}{(x+1)(x+2)}$

Showing the 12 most recent of 33 on this concept.

1. 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

2. 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).

3. 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 $\boxed{\frac{5}{2}}$.

4. 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).

5. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If $\dfrac{3x+2}{(x+1)(2x^2+3)} = \dfrac{A}{x+1} + \dfrac{Bx+C}{2x^2+3}$, then $A-B+C=$ (A) 2 (B) 1 (C) 3 (D) 6

   ›Reveal solutionSolution

   We find the constants in the partial fraction decomposition by clearing denominators and equating coefficients, then compute $A - B + C = 1$.

   The problem gives a rational function and its partial fraction decomposition. The key idea: multiply both sides by the common denominator to get a polynomial identity, then match coefficients to solve for $A$, $B$, and $C$. Once we have them, the expression $A - B + C$ is straightforward.

   ---

   1. Set up the equation We have

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

   Multiply both sides by $(x+1)(2x^2+3)$ to clear denominators:

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

   2. Expand the right-hand side First term: $A(2x^2+3) = 2A x^2 + 3A$. Second term: $(Bx+C)(x+1) = Bx^2 + Bx + Cx + C = Bx^2 + (B+C)x + C$. Adding them:

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

   3. Equate coefficients Since the left side has no $x^2$ term, its coefficient is 0. So:

   $\begin{cases} \text{Coefficient of } x^2: & 2A + B = 0 \quad (1) \\ \text{Coefficient of } x: & B + C = 3 \quad (2) \\ \text{Constant term:} & 3A + C = 2 \quad (3) \end{cases}$

   4. Solve the system

      From (1): $B = -2A$.

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

      Substitute into (3): $3A + (3 + 2A) = 2 \Rightarrow 5A + 3 = 2 \Rightarrow 5A = -1 \Rightarrow A = -\frac{1}{5}$.

      Then $B = -2\left(-\frac{1}{5}\right) = \frac{2}{5}$, and $C = 3 + 2\left(-\frac{1}{5}\right) = 3 - \frac{2}{5} = \frac{13}{5}$.

   5. Compute $A - B + C$

   $A - B + C = -\frac{1}{5} - \frac{2}{5} + \frac{13}{5} = \frac{-1 - 2 + 13}{5} = \frac{10}{5} = 2.$

   Watch out

   A common mistake is forgetting that the numerator of the second fraction is $Bx + C$, not just a constant. Also, when equating coefficients, note that the left side has no $x^2$ term — that gives $2A + B = 0$, not something else.

   Tip

   You can also solve by plugging in convenient $x$ values (like $x = -1$ to get $A$ directly), but the coefficient method is systematic and avoids fractions until the end.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

6. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $\dfrac{x+1}{(x^2+1)(x-1)^2} = \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-1} + \dfrac{D}{(x-1)^2}$, then $A+B+C+D=$ (A) $-\dfrac{1}{2}$ (B) $\dfrac{1}{2}$ (C) $1$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   We decompose the given rational function into partial fractions by equating numerators and solving for the coefficients $A, B, C, D$. The sum $A+B+C+D$ is $\boxed{\dfrac{1}{2}}$.

   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 a sum of simpler fractions. This process is particularly useful in calculus for integration, but it's also a fundamental algebraic skill.

   The core idea is that any proper rational function $\frac{P(x)}{Q(x)}$ (where the degree of $P(x)$ is less than the degree of $Q(x)$) can be expressed as a sum of simpler fractions whose denominators are the factors of $Q(x)$. The form of these simpler fractions depends on the nature of the factors in the denominator $Q(x)$:

   1. Linear Factor $(ax+b)$: For each non-repeated linear factor, there is a term of the form $\frac{A}{ax+b}$.
   2. Repeated Linear Factor $(ax+b)^n$: For each repeated linear factor, there are $n$ terms of the form $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}$.
   3. Irreducible Quadratic Factor $(ax^2+bx+c)$: For each non-repeated irreducible quadratic factor (one that cannot be factored into linear factors with real coefficients, i.e., $b^2-4ac < 0$), there is a term of the form $\frac{Ax+B}{ax^2+bx+c}$.
   4. Repeated Irreducible Quadratic Factor $(ax^2+bx+c)^n$: For each repeated irreducible quadratic factor, there are $n$ terms of the form $\frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + \dots + \frac{A_nx+B_n}{(ax^2+bx+c)^n}$.

   In this problem, the denominator is $(x^2+1)(x-1)^2$.

   • $(x^2+1)$ is an irreducible quadratic factor.
   • $(x-1)^2$ is a repeated linear factor.

   The given partial fraction form $\dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-1} + \dfrac{D}{(x-1)^2}$ correctly follows these rules. Our task is to find the unknown coefficients $A, B, C, D$.

   Step-by-Step Solution

   1. Combine the terms on the right-hand side: To find the coefficients, we first combine the partial fractions on the right-hand side using a common denominator, which will be $(x^2+1)(x-1)^2$.

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

   2. Equate the numerators: Since the denominators are identical, the numerators must be equal.

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

   This equation must hold true for all values of $x$. We can use a combination of substituting convenient values of $x$ and equating coefficients of powers of $x$ to find $A, B, C, D$.
   

   3. Find the coefficients using substitution and equating coefficients:

   *   **Substitute $x=1$:** This value makes the terms involving $(x-1)$ and $(x-1)^2$ zero, allowing us to find $D$ directly.
   

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

   $2 = (A+B)(0) + C(2)(0) + D(2)$

   $2 = 2D \implies \mathbf{D=1}$

   *   **Substitute $x=0$:** This value often simplifies expressions involving $x$.
       Substitute $D=1$ into the main numerator equation:
   

   $x+1 = (Ax+B)(x-1)^2 + C(x^2+1)(x-1) + (x^2+1)$

       Now, substitute $x=0$:
   

   $0+1 = (A(0)+B)(0-1)^2 + C(0^2+1)(0-1) + (0^2+1)$

   $1 = B(1)^2 + C(1)(-1) + 1$

   $1 = B - C + 1$

   $0 = B - C \implies \mathbf{B=C}$

   *   **Substitute $x=-1$:** This is another convenient value.
       Substitute $D=1$ and $B=C$ into the main numerator equation:
   

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

   $0 = (-A+B)(-2)^2 + C(1+1)(-2) + (1+1)$

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

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

       Since $B=C$, we can substitute $C$ with $B$:
   

   $0 = 4(-A+B) - 4B + 2$

   $0 = -4A + 4B - 4B + 2$

   $0 = -4A + 2$

   $4A = 2 \implies \mathbf{A=\frac{1}{2}}$

   *   **Find $B$ and $C$:**
       We found $A=\frac{1}{2}$ and $B=C$. We need one more piece of information. Let's equate the coefficients of $x^3$ from both sides of the numerator equation.
       The left side is $x+1$, so the coefficient of $x^3$ is $0$.
       For the right side, let's expand the terms that produce $x^3$:
   

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

   $C(x^2+1)(x-1) = C(x^3-x^2+x-1) = Cx^3 - Cx^2 + Cx - C$

   $D(x^2+1) = Dx^2+D$

       The coefficient of $x^3$ on the right side is $A+C$.
       Equating coefficients of $x^3$:
   

   $A+C = 0$

       Since $A=\frac{1}{2}$:
   

   $\frac{1}{2} + C = 0 \implies \mathbf{C=-\frac{1}{2}}$

       And since $B=C$:
   

   $\mathbf{B=-\frac{1}{2}}$

   *   **Summary of coefficients:**
       $A = \frac{1}{2}$
       $B = -\frac{1}{2}$
       $C = -\frac{1}{2}$
       $D = 1$
   

   4. Calculate $A+B+C+D$:

   $A+B+C+D = \frac{1}{2} + \left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right) + 1$

   $A+B+C+D = 0 - \frac{1}{2} + 1$

   $A+B+C+D = \frac{1}{2}$

   ✓Final answer

   The value of $A+B+C+D$ is $\boxed{\dfrac{1}{2}}$.

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

   ›Reveal solutionSolution

   Clearing denominators and matching coefficients gives $A=0,\ B=2,\ C=0,\ D=-1$, so $A+B+C+D=1$ — option (D).

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

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

   Expand and collect powers of $x$:

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

   Match coefficients:

   • $x^3:\ A+C=0$
   • $x^2:\ B+D=1$
   • $x^1:\ 2A+C=0$
   • $x^0:\ 2B+D=3$

   Solve. Subtracting $A+C=0$ from $2A+C=0$ gives $A=0$, hence $C=0$. Subtracting $B+D=1$ from $2B+D=3$ gives $B=2$, hence $D=-1$.

   Sum: $A+B+C+D=0+2+0-1=1.$

   ✓Final answer

   $A+B+C+D=1$ — option (D).

8. 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

9. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $\dfrac{x^2+7}{(x^2+1)(x-2)} = \dfrac{A}{x-2} + \dfrac{Bx+C}{x^2+1}$, then the determinant of the matrix $\begin{pmatrix} A & B \\ C & \dfrac{2}{5} \end{pmatrix}$ is (A) $5$ (B) $-5$ (C) $\dfrac{94}{25}$ (D) $-2$

   ›Reveal solutionSolution

   We decompose the given rational function into partial fractions, solve for $A$, $B$, and $C$, then compute the determinant of the $2\times2$ matrix formed by them. The determinant is $\boxed{-5}$.

   The problem gives a partial fraction decomposition and asks for the determinant of a matrix built from the coefficients. The key is to find $A$, $B$, and $C$ correctly — then the determinant is just a quick calculation.

   We start with the identity:

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

   This holds for all $x$ (except where denominators vanish). Multiply both sides by $(x^2+1)(x-2)$ to clear denominators:

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

   1. Expand the right-hand side

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

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

   Adding them:

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

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

   $\begin{cases} A + B = 1 & \text{(coefficient of $x^2$)} \\ C - 2B = 0 & \text{(coefficient of $x$)} \\ A - 2C = 7 & \text{(constant term)} \end{cases}$

   3. Solve the system From the second equation: $C = 2B$. Substitute into the third: $A - 2(2B) = A - 4B = 7$. From the first: $A = 1 - B$. Plug into $A - 4B = 7$:

   $(1 - B) - 4B = 7 \quad\Rightarrow\quad 1 - 5B = 7 \quad\Rightarrow\quad -5B = 6 \quad\Rightarrow\quad B = -\frac{6}{5}$

   Then $A = 1 - (-\frac{6}{5}) = 1 + \frac{6}{5} = \frac{11}{5}$.

   And $C = 2B = 2 \cdot (-\frac{6}{5}) = -\frac{12}{5}$.

   So we have:

   $A = \frac{11}{5},\quad B = -\frac{6}{5},\quad C = -\frac{12}{5}$

   4. Form the matrix and compute the determinant The matrix is:

   $\begin{pmatrix} A & B \\ C & \dfrac{2}{5} \end{pmatrix} = \begin{pmatrix} \frac{11}{5} & -\frac{6}{5} \\[4pt] -\frac{12}{5} & \frac{2}{5} \end{pmatrix}$

   Its determinant is:

   $\det = \left(\frac{11}{5}\right)\left(\frac{2}{5}\right) - \left(-\frac{6}{5}\right)\left(-\frac{12}{5}\right) = \frac{22}{25} - \frac{72}{25} = -\frac{50}{25} = -2$

   Watch out

   A common mistake is to forget the minus sign when expanding $(Bx+C)(x-2)$ — the term $-2Bx$ is easy to miss. Also, when computing the determinant, be careful with the signs of the off-diagonal product: $B$ and $C$ are both negative, so their product is positive, but the formula subtracts it.

   ✓Final answer

   The determinant is $\boxed{-2}$, which corresponds to option (D).

10. 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).

11. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.If $\dfrac{x^2 - x + 1}{(x^2 + 1)(x^2 + x + 1)} = \dfrac{Ax + B}{x^2 + 1} + \dfrac{Cx + D}{x^2 + x + 1}$ then $A + 2B + C + 2D =$ (A) $0$ (B) $1$ (C) $-1$ (D) $2$

    ›Reveal solutionSolution

    This is a partial fractions problem where we match coefficients after clearing denominators. The value of $A+2B+C+2D$ is $0$.

    The core idea here is that when you have a rational expression and you decompose it into partial fractions, the numerators on the right-hand side are linear (since the denominators are irreducible quadratics). To find the constants, you multiply through by the common denominator and then compare coefficients of powers of $x$. That gives a system of linear equations. Once you solve for $A, B, C, D$, you just plug into the required combination.

    Let’s work through it step by step.

    1. Set up the equation We are given:

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

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

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

    2. Expand both products First term:

    $(Ax+B)(x^2+x+1) = A x^3 + A x^2 + A x + B x^2 + B x + B$

    That is:

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

    Second term:

    $(Cx+D)(x^2+1) = C x^3 + C x + D x^2 + D$

    That is:

    $C x^3 + D x^2 + C x + D$

    3. Add them together

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

    4. Compare coefficients Since the left side has no $x^3$ term, its coefficient is $0$. So:

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

    5. Solve the system

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

       From (4): $D = 1 - B$.

       Substitute into (2): $A + B + (1 - B) = 1 \implies A + 1 = 1 \implies A = 0$.

       Then $C = 0$ from (1).

       Now (3): $0 + B + 0 = -1 \implies B = -1$.

       Then from (4): $D = 1 - (-1) = 2$.

       So we have:

    $A = 0,\quad B = -1,\quad C = 0,\quad D = 2$

    6. Compute the required expression

    $A + 2B + C + 2D = 0 + 2(-1) + 0 + 2(2) = -2 + 4 = 2$

    Watch out

    A common mistake is to forget the constant term when expanding $(Cx+D)(x^2+1)$ — the $D$ multiplies the $1$ to give $D$, not $Dx^2$. Also, be careful with signs when comparing coefficients: the left side is $x^2 - x + 1$, so the coefficient of $x$ is $-1$, not $+1$.

    ✓Final answer

    The value is $\boxed{2}$, which corresponds to option (D).

12. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.If $\dfrac{x^2+1}{(x^2+2)(x^2+3)} = \dfrac{Ax+B}{x^2+2} + \dfrac{Cx+D}{x^2+3}$, then $A+B+C+D=$ (A) $0$ (B) $1$ (C) $-1$ (D) $6$

    ›Reveal solutionSolution

    We decompose the rational function into partial fractions with linear numerators, then equate coefficients to solve for A, B, C, D; summing them gives 0.

    We are given:

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

    and need $A+B+C+D$.

    Concept & Intuition

    Since the denominator factors are irreducible quadratics, each partial fraction gets a linear numerator ($Ax+B$ form). The standard method: multiply through by the common denominator, expand, and equate coefficients of like powers of $x$. This yields a system of equations for $A,B,C,D$. Summing them is then trivial.

    Step-by-step

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

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

    2. Expand each term

    $(Ax+B)(x^2+3) = Ax^3 + 3Ax + Bx^2 + 3B$

    $(Cx+D)(x^2+2) = Cx^3 + 2Cx + Dx^2 + 2D$

    3. Combine like powers

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

    4. Equate coefficients

       • Coefficient of $x^3$: $A + C = 0$
       • Coefficient of $x^2$: $B + D = 1$
       • Coefficient of $x^1$: $3A + 2C = 0$
       • Constant term: $3B + 2D = 1$

    5. Solve the system

       From $A+C=0$ we have $C = -A$.

       Substitute into $3A+2C=0$: $3A + 2(-A) = A = 0$ → $A=0$, then $C=0$.

       From $B+D=1$ we have $D = 1-B$.

       Substitute into $3B+2D=1$: $3B + 2(1-B) = 3B + 2 - 2B = B+2 = 1$ → $B = -1$, then $D = 1-(-1)=2$.

    6. Sum the constants

    $A+B+C+D = 0 + (-1) + 0 + 2 = 1$

    Watch out

    A common mistake is forgetting that the numerators for irreducible quadratics must be linear ($Ax+B$), not just constants. Using constants would give a wrong system.

    Tip

    Notice that the $x^3$ and $x$ coefficients forced $A=C=0$ immediately, so the sum reduces to $B+D$, which from the $x^2$ coefficient is already 1 — no need to solve fully!

    ✓Final answer

    The correct option is (B).

    ANSWER: B