Q.Integrate the following function: $\frac{x}{(x - 1)(x - 2)(x - 3)}$

Showing the 12 most recent of 33 on this concept.

1. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If $\dfrac{1}{(x-1)(x-2)(x-3)} = \dfrac{A}{x-1} + \dfrac{B}{x-2} + \dfrac{C}{x-3}$, and $\dfrac{x}{(x-1)(x-2)(x-3)} = \dfrac{P}{x-1} + \dfrac{Q}{x-2} + \dfrac{R}{x-3}$ then $A + 2B + 3C =$ (A) $P + Q + R$ (B) $P + 2Q + 3R$ (C) $3P + 2Q + R$ (D) $AP + BQ + CR$

   ›Reveal solutionSolution

   The key idea is to find the partial fraction constants by comparing numerators after clearing denominators, then compute $A+2B+3C$ and match it to a combination of $P, Q, R$. The result is $A+2B+3C = P+2Q+3R$, so option (B) is correct.

   The problem gives two partial fraction expansions for very similar rational functions. The first has numerator $1$, the second has numerator $x$. The constants $A, B, C$ and $P, Q, R$ are determined uniquely by the denominators. The question asks for a linear combination of $A, B, C$ and wants it expressed in terms of $P, Q, R$. The natural approach is to compute each constant explicitly, then evaluate the required sum.

   1. Find $A, B, C$. Start with

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

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

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

   This identity holds for all $x$. To isolate each constant, substitute the roots of the denominators.

   • For $x=1$: the terms with $B$ and $C$ vanish because $(x-1)$ factors become zero.

   $1 = A(1-2)(1-3) = A(-1)(-2) = 2A \quad\Rightarrow\quad A = \frac12.$

   • For $x=2$:

   $1 = B(2-1)(2-3) = B(1)(-1) = -B \quad\Rightarrow\quad B = -1.$

   • For $x=3$:

   $1 = C(3-1)(3-2) = C(2)(1) = 2C \quad\Rightarrow\quad C = \frac12.$

   So $A = \frac12$, $B = -1$, $C = \frac12$.

   2. Find $P, Q, R$. Now

   $\frac{x}{(x-1)(x-2)(x-3)} = \frac{P}{x-1} + \frac{Q}{x-2} + \frac{R}{x-3}.$

   Multiply through by the denominator:

   $x = P(x-2)(x-3) + Q(x-1)(x-3) + R(x-1)(x-2).$

   Again substitute the roots.

   • For $x=1$:

   $1 = P(1-2)(1-3) = P(-1)(-2) = 2P \quad\Rightarrow\quad P = \frac12.$

   • For $x=2$:

   $2 = Q(2-1)(2-3) = Q(1)(-1) = -Q \quad\Rightarrow\quad Q = -2.$

   • For $x=3$:

   $3 = R(3-1)(3-2) = R(2)(1) = 2R \quad\Rightarrow\quad R = \frac32.$

   So $P = \frac12$, $Q = -2$, $R = \frac32$.

   3. Compute $A + 2B + 3C$. $$$$

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

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3. 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:

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

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

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5. 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 $B = -A = \frac{17}{5}$. …

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6. 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:

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

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7. 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) …

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8. 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:

      • $x^6$: $B + C + D = 0$
      • $x^4$: $A + 5B + 3C + 2D = 0$ …

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9. 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  
    $$ …
   

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10. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.If $\dfrac{x-2}{x^2(2x-3)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{2x-3}$ then $2(A-C)=$ (A) 3B (B) 2B (C) 0 (D) B

    ›Reveal solutionSolution

    To find the coefficients A, B, and C in the partial fraction decomposition, we equate the numerators after combining the terms on the right-hand side. By substituting specific values of $x$ or comparing coefficients, we find $A = 1/9$, $B = 2/3$, and $C = -2/9$. The expression $2(A-C)$ then evaluates to $2/3$, which is equal to $B$.

    Partial fraction decomposition is a technique used to break down a complex rational function into a sum of simpler fractions. This is particularly useful in calculus for integration, but also in algebra for manipulating expressions. The core idea is that any proper rational function (where the degree of the numerator is less than the degree of the denominator) can be expressed as a sum of fractions whose denominators are the factors of the original denominator.

    When the denominator has repeated linear factors, like $x^2$ in this problem, the decomposition must include a term for each power of the factor up to its multiplicity. For a factor $(ax+b)^n$, we include terms $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}$. For distinct linear factors, like $(2x-3)$, we simply have a term $\frac{C}{2x-3}$.

    The strategy is to combine the partial fractions on the right-hand side, equate the resulting numerator to the original numerator, and then solve for the unknown coefficients (A, B, C) by either substituting convenient values of $x$ or by comparing coefficients of like powers of $x$.

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

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

    To find the coefficients $A$, $B$, and $C$, we first combine the terms on the right-hand side by finding a common denominator, which is $x^2(2x-3)$.
    

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

    Since the denominators are now identical, the numerators must be equal:
    

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

    Expand the right-hand side:
    

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

    Rearrange the terms by powers of $x$:
    

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

    2. Determine the coefficients using strategic substitution and comparison:

       We can find the coefficients by substituting values of $x$ that make certain terms zero, or by comparing the coefficients of $x^2$, $x$, and the constant term on both sides of the equation.

       • Find B: Substitute $x=0$ into the equation $x-2 = A(x)(2x-3) + B(2x-3) + C(x^2)$. This eliminates the terms with $A$ and $C$:

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

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

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

    *   **Find C:** Substitute $x=\frac{3}{2}$ (which makes $2x-3=0$) into the equation $x-2 = A(x)(2x-3) + B(2x-3) + C(x^2)$. This eliminates the terms with $A$ and $B$:
    

    $\frac{3}{2}-2 = A\left(\frac{3}{2}\right)(0) + B(0) + C\left(\frac{3}{2}\right)^2$

    $\frac{3-4}{2} = 0 + 0 + C\left(\frac{9}{4}\right)$

    $-\frac{1}{2} = \frac{9}{4}C \implies C = -\frac{1}{2} \times \frac{4}{9} = -\frac{2}{9}$

    *   **Find A:** Now that we have $B$ and $C$, we can find $A$ by comparing the coefficients of $x^2$ from the expanded equation: …
    

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11. 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): $C = 2A$. …

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 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) $1$ (B) $0$ (C) $-1$ (D) $6$

    ›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 $A+B+C+D$ turns out to be $0$.

    We are given the partial fraction decomposition:

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

    We need $A+B+C+D$. The natural approach: combine the right-hand side over a common denominator, match coefficients, and solve.

    ---

    1. Combine the fractions on the right.

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

    Since the denominators are already equal, we equate the numerators:

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

    ---

    2. Expand both products.

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

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

    Add them:

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

    So we have:

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

    ---

    3. Equate coefficients.

    The left side has no $x^3$ term, no $x$ term, constant term $1$, and $x^2$ coefficient $1$. So:

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

    ---

    4. Solve the system.

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

    Plug into $3A + 2C = 0$: $3A + 2(-A) = 3A - 2A = A = 0$.

    Thus $A = 0$ and then $C = 0$. …

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