Q.Evaluate the definite integral: $\int_{1}^{2} \frac{5x^2}{x^2+4x+3} \, dx$

Showing the 12 most recent of 33 on this concept.

1. 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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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 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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4. 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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5. 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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6. 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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7. 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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8. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $\dfrac{x^2 - 3x + 2}{(x-4)(x-3)^2} = \dfrac{A}{x-4} + \dfrac{B}{x-3} + \dfrac{C}{(x-3)^2}$ then $A+B+C=$ (A) $1$ (B) $0$ (C) $-1$ (D) $5$

   ›Reveal solutionSolution

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

   We are given the partial fraction decomposition:

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

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

   ---

   Concept & Intuition

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

   ---

   Step-by-step solution

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

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

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

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

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

   On the right-hand side:

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

   Simplify:

   $= A + 2B - 2C$

   So we have:

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

   3. Find another relation

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

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

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

      So:

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

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

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

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

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

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

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

   $\Rightarrow 7A + 8B = 2$

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

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

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

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

      Set $x=1$:

      LHS: $1 - 3 + 2 = 0$

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

      So:

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

   Now we have three equations:

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

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

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

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9. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.If $\dfrac{6x^4 + 13x^3 + 2x^2 - x + 3}{2x^2 + 3x - 2} = f(x) + \dfrac{A}{ax-1} + \dfrac{B}{x+b}$, then $f(1) + a \cdot B + b \cdot A =$ (A) 8 (B) 12 (C) 4 (D) 6

   ›Reveal solutionSolution

   Long division gives $f(x)=3x^2+2x+1$ with $a=2,b=2,A=2,B=-1$, so $f(1)+aB+bA = 6-2+4 = 8$ (option A).

   Factor the denominator: $2x^2+3x-2 = (2x-1)(x+2)$, so $ax-1 = 2x-1 \Rightarrow a=2$ and $x+b = x+2 \Rightarrow b=2$.

   Polynomial division of $6x^4+13x^3+2x^2-x+3$ by $2x^2+3x-2$:

   $f(x) = 3x^2 + 2x + 1, \qquad \text{remainder } = 5.$

   Partial fractions of the remainder:

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

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10. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.If $\dfrac{32x^2 + 186x}{(x^2+1)(x+5)} = \dfrac{37x+1}{x^2+1} + \dfrac{\lambda}{x+5}$, then $\dfrac{\lambda}{2} =$ (A) $-5$ (B) $-\dfrac{7}{2}$ (C) $-\dfrac{3}{2}$ (D) $-\dfrac{5}{2}$

    ›Reveal solutionSolution

    To find $\lambda$, we combine the fractions on the right side, equate the numerators, and then substitute a strategic value for $x$ to solve for $\lambda$. The value of $\lambda$ is $-5$, so $\frac{\lambda}{2}$ is $\boxed{-\frac{5}{2}}$.

    The problem asks us to find the value of $\frac{\lambda}{2}$ given an identity involving partial fractions. The core idea here is that if two rational expressions are equal, and their denominators are identical, then their numerators must also be identical. This allows us to form an equation involving $x$ and $\lambda$, which we can then solve for $\lambda$.

    There are two main approaches to solve for the unknown constants in partial fraction decomposition:

    1. Equating Coefficients: Expand the right-hand side, collect terms by powers of $x$, and then equate the coefficients of corresponding powers of $x$ on both sides of the equation. This will give a system of linear equations to solve for the constants.
    2. Substitution Method: Choose specific values of $x$ that simplify the equation, often by making certain terms zero. This is particularly effective when dealing with linear factors in the denominator.

    In this problem, the substitution method is more direct for finding $\lambda$.

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

    1. Combine the fractions on the right-hand side: The given equation is:

    $\frac{32x^2 + 186x}{(x^2+1)(x+5)} = \frac{37x+1}{x^2+1} + \frac{\lambda}{x+5}$

    To combine the terms on the right-hand side, we find a common denominator, which is $(x^2+1)(x+5)$:
    

    $\frac{37x+1}{x^2+1} + \frac{\lambda}{x+5} = \frac{(37x+1)(x+5)}{(x^2+1)(x+5)} + \frac{\lambda(x^2+1)}{(x+5)(x^2+1)}$

    $= \frac{(37x+1)(x+5) + \lambda(x^2+1)}{(x^2+1)(x+5)}$

    2. Equate the numerators: Since the denominators on both sides of the original equation are now identical, the numerators must be equal for the identity to hold true for all values of $x$:

    $32x^2 + 186x = (37x+1)(x+5) + \lambda(x^2+1)$

    3. Solve for $\lambda$ using a strategic substitution: …

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11. 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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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 $D = 1-B$. …

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