Q.Prove that $\int_{1}^{3}\frac{dx}{x^2(x+1)}=\frac{2}{3}+\log\frac{2}{3}$

Showing the 12 most recent of 33 on this concept.

1. 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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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:

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

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

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4. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If $\dfrac{3x+1}{(x-1)^2(x^2+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$ (A) $-1$ (B) $2$ (C) $0$ (D) $1$

   ›Reveal solutionSolution

   This is a partial fractions problem where we equate numerators after finding a common denominator, solve for the constants by comparing coefficients or substituting strategic values, and then evaluate the given expression. The final value is $0$.

   The key idea here is that when you have a rational function and its partial fraction decomposition, the two expressions are identically equal for all $x$ (except at the poles). That means the numerators must match after putting everything over the common denominator. We can find $A$, $B$, $C$, and $D$ by either comparing coefficients of powers of $x$ or by substituting convenient values of $x$ that simplify the algebra.

   Let’s work through it step by step.

   1. Set up the equation. We are given:

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

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

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

   2. Expand each term carefully.

      • First term: $A(x-1)(x^2+1) = A[(x)(x^2+1) - 1\cdot(x^2+1)] = A(x^3 + x - x^2 - 1) = A(x^3 - x^2 + x - 1)$
      • Second term: $B(x^2+1) = Bx^2 + B$
      • Third term: $(Cx+D)(x-1)^2 = (Cx+D)(x^2 - 2x + 1)$ Expand: $Cx(x^2-2x+1) = Cx^3 - 2Cx^2 + Cx$ and $D(x^2-2x+1) = Dx^2 - 2Dx + D$ So together: $Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$

   3. Combine all terms on the right-hand side.

      Collecting by powers of $x$:

   $\begin{aligned} x^3 &: A + C \\ x^2 &: -A + B - 2C + D \\ x^1 &: A + C - 2D \\ x^0 &: -A + B + D \end{aligned}$

   The left-hand side is $3x+1$, which is $0\cdot x^3 + 0\cdot x^2 + 3x + 1$.

   4. Equate coefficients. This gives us a system of equations:

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

   5. Solve the system. From (1): $C = -A$. Substitute into (3): $A + (-A) - 2D = 3 \implies -2D = 3 \implies D = -\frac{3}{2}$. Now (4): $-A + B - \frac{3}{2} = 1 \implies -A + B = \frac{5}{2}$. And (2): $-A + B - 2(-A) - \frac{3}{2} = 0 \implies -A + B + 2A - \frac{3}{2} = 0 \implies A + B = \frac{3}{2}$. …

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5. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If $\dfrac{3x+1}{(x-1)^2(x^2+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$ (A) $0$ (B) $1$ (C) $2$ (D) $-1$

   ›Reveal solutionSolution

   We find the constants in the partial fraction decomposition by clearing denominators and equating coefficients, then compute $2(A-C+B+D)$ to get $-1$, which corresponds to option (D).

   We are given the partial fraction decomposition:

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

   Our goal is to find $A, B, C, D$ and then evaluate $2(A - C + B + D)$.

   ---

   Concept & Intuition

   Partial fractions let us break a complicated rational expression into simpler pieces. The denominators here are $(x-1)$, $(x-1)^2$, and the irreducible quadratic $x^2+1$. To find the unknown constants, we multiply both sides by the common denominator $(x-1)^2(x^2+1)$, which gives a polynomial identity. Then we compare coefficients of like powers of $x$ (or substitute convenient values of $x$) to solve for $A, B, C, D$.

   ---

   Step-by-step solution

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

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

   2. Expand each term

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

      Expand the last one carefully:

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

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

   3. Sum all contributions

      Collecting like powers from all three pieces:

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

      The left side is $3x + 1$, which we write as $0x^3 + 0x^2 + 3x + 1$.

   4. Set up the system of equations

      Equating coefficients:

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

   5. Solve the system

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

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

      Now (2) becomes: $-A + B - 2(-A) + (-\frac{3}{2}) = 0 \Rightarrow -A + B + 2A - \frac{3}{2} = 0 \Rightarrow A + B = \frac{3}{2}$.

      Equation (4): $-A + B - \frac{3}{2} = 1 \Rightarrow -A + B = \frac{5}{2}$.

      Now solve the two equations in $A, B$:

      $$$$

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6. 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:

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

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7. 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 $D = 1-B$. …

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8. 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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9. 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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10. 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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11. 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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12. 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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