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

Showing the 12 most recent of 33 on this concept.

1. 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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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 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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4. 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: …
   

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5. 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 $C=0$. …

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

   ›Reveal solutionSolution

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

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

   1. Set up the equation We have

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

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

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

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

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

   So:

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

   3. Expand each term

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

   4. Collect coefficients

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

      The right side, collecting powers:

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

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

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

   5. Solve the system From (1): $C = 2A$. …

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10. 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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11. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.If $\dfrac{x^4}{(x-1)(x-2)(x-3)} = p(x) + \dfrac{A}{x-1} + \dfrac{B}{x-2} + \dfrac{C}{x-3}$, then $p\left(\dfrac{3}{2}\right) + C =$ (A) $0$ (B) $8$ (C) $\dfrac{17}{2}$ (D) $48$

    ›Reveal solutionSolution

    The key idea is to perform polynomial long division on the rational function, then use the Heaviside cover-up method for partial fractions. The final result is that $p(3/2) + C = 8$, so the correct option is (B).

    We start with the given identity:

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

    Here $p(x)$ is the quotient polynomial from dividing $x^4$ by $(x-1)(x-2)(x-3)$, and $A, B, C$ are constants from the partial fraction decomposition of the remainder.

    ---

    1. Find the degree of $p(x)$

    The denominator $(x-1)(x-2)(x-3)$ is degree 3. The numerator $x^4$ is degree 4.

    Since degree(numerator) = degree(denominator) + 1, the quotient $p(x)$ will be a linear polynomial:

    $p(x) = ax + b$

    ---

    2. Perform polynomial long division conceptually

    Instead of doing the full division, note that:

    $\frac{x^4}{(x-1)(x-2)(x-3)} = \frac{x^4}{x^3 - 6x^2 + 11x - 6}$

    Divide:

    • $x^4 \div x^3 = x$, so first term is $x$.
    • Multiply back: $x(x^3 - 6x^2 + 11x - 6) = x^4 - 6x^3 + 11x^2 - 6x$.
    • Subtract from $x^4$: remainder is $6x^3 - 11x^2 + 6x$.
    • Now divide $6x^3$ by $x^3$ gives $6$.
    • Multiply: $6(x^3 - 6x^2 + 11x - 6) = 6x^3 - 36x^2 + 66x - 36$.
    • Subtract: $(6x^3 - 11x^2 + 6x) - (6x^3 - 36x^2 + 66x - 36) = 25x^2 - 60x + 36$.

    Thus:

    $\frac{x^4}{(x-1)(x-2)(x-3)} = x + 6 + \frac{25x^2 - 60x + 36}{(x-1)(x-2)(x-3)}$

    So $p(x) = x + 6$.

    ---

    3. Find $C$ using the cover-up method

    We have:

    $\frac{25x^2 - 60x + 36}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$

    To find $C$, multiply both sides by $(x-3)$ and then set $x = 3$:

    $$$$

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12. 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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