Q.Integrate the following function: $\frac{5x}{(x + 1)(x^2 - 4)}$

Showing the 12 most recent of 63 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x}{x^2 - 5x + 4}\,dx =$ (A) $\dfrac{1}{3}\log\dfrac{(x-4)^4}{|x-1|} + c$ (B) $\dfrac{4}{3}\log\dfrac{|x-4|}{(x-1)^4} + c$ (C) $-\dfrac{1}{3}\log\dfrac{(x-4)^2}{|x-1|}$ (D) $-\dfrac{4}{3}\log\dfrac{|x-4|}{(x-1)^4} + c$

   ›Reveal solutionSolution

   Partial fraction decomposition of a rational function with distinct linear factors, followed by direct log integration and recombination, gives $\dfrac13\log\dfrac{(x-4)^4}{|x-1|}+c$.

   Concept and Intuition

   Whenever the denominator of a rational integrand factors into distinct linear terms, partial fractions break it into simpler pieces, each of which integrates to a logarithm. Combining the two resulting logarithm terms back into a single log-of-a-ratio (using log rules $a\log m - b\log n = \log\frac{m^a}{n^b}$) is what makes the answer match a compact multiple-choice form.

   Step-by-Step Solution

   1. Factor the denominator: $x^2-5x+4 = (x-1)(x-4)$.
   2. Write $\dfrac{x}{(x-1)(x-4)} = \dfrac{A}{x-1}+\dfrac{B}{x-4}$.
   3. Multiply through: $x = A(x-4)+B(x-1)$.
   4. Set $x=1$: $1 = A(1-4) = -3A \Rightarrow A = -\dfrac13$.
   5. Set $x=4$: $4 = B(4-1) = 3B \Rightarrow B = \dfrac43$.
   6. So the integral is $\displaystyle\int\left(-\dfrac{1/3}{x-1}+\dfrac{4/3}{x-4}\right)dx = -\dfrac13\log|x-1| + \dfrac43\log|x-4| + c$. …

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2. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.The partial fraction of $\dfrac{x^2}{x^2 + 3x - 4}$ is ________ (A) $1 + \dfrac{-16}{5(x+4)} + \dfrac{1}{5(x-1)}$ (B) $1 + \dfrac{-1}{x+4} + \dfrac{1}{x-1}$ (C) $1 + \dfrac{-13}{5(x+4)} + \dfrac{1}{5(x-1)}$ (D) $\dfrac{2}{x+4} + \dfrac{1}{x-1}$

   ›Reveal solutionSolution

   Since the numerator's degree equals the denominator's degree, perform polynomial division first, then resolve the remaining proper fraction into partial fractions. Answer: (A).

   Concept and Intuition

   Partial fraction decomposition applies to a proper rational function (numerator degree less than denominator degree). Here both are degree 2, so we must first extract the constant (integer) part via division, leaving a proper fraction to decompose.

   Step-by-Step Solution

   1. Factor the denominator: $x^2+3x-4=(x+4)(x-1)$.
   2. Divide: $x^2 = (x^2+3x-4) - (3x-4)$, so $\dfrac{x^2}{x^2+3x-4} = 1 - \dfrac{3x-4}{(x+4)(x-1)}$.
   3. Decompose $\dfrac{3x-4}{(x+4)(x-1)} = \dfrac{A}{x+4}+\dfrac{B}{x-1}$, so $3x-4=A(x-1)+B(x+4)$.
   4. Set $x=1$: $-1 = 5B \Rightarrow B=-\dfrac15$.
   5. Set $x=-4$: $-16 = -5A \Rightarrow A=\dfrac{16}{5}$.
   6. So $\dfrac{3x-4}{(x+4)(x-1)} = \dfrac{16}{5(x+4)} - \dfrac{1}{5(x-1)}$. …

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3. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Which of the following is a partial fraction of $\dfrac{-x^{2}+6x+13}{(3x+5)(x^{2}+4x+4)} =$ (A) $\dfrac{3}{3x+5} + \dfrac{-1}{x+2} + \dfrac{2}{(x+2)^{2}}$ (B) $\dfrac{2}{3x+5} + \dfrac{-1}{x+2} + \dfrac{3}{(x+2)^{2}}$ (C) $\dfrac{-1}{3x+5} + \dfrac{2}{x+2} + \dfrac{3}{(x+2)^{2}}$ (D) $\dfrac{3}{3x+5} + \dfrac{2}{x+2} + \dfrac{-1}{(x+2)^{2}}$

   ›Reveal solutionSolution

   Clearing denominators and plugging in the repeated-root value, the linear-factor root, and matching one coefficient gives $A=2,\ B=-1,\ C=3$.

   Concept and Intuition

   For a repeated linear factor, plugging in its root isolates the coefficient of the highest power of that factor directly; the simple linear factor's root isolates its own coefficient; any remaining coefficient is found by matching a convenient power of $x$.

   Step-by-Step Solution

   1. Write $-x^2+6x+13 = A(x+2)^2 + B(3x+5)(x+2) + C(3x+5)$.
   2. At $x=-2$: LHS $=-4-12+13=-3$; RHS $=C(3(-2)+5)=-C$. So $C=3$.
   3. At $x=-5/3$: LHS $=-25/9-10+13=2/9$; RHS $=A(x+2)^2=A(1/3)^2=A/9$. So $A=2$. …

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4. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x^2}{(x^2-1)(x^2+1)}dx=$ (A) $\dfrac{1}{4}\log\left|\dfrac{x+1}{x-1}\right|-\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$ (B) $\dfrac{1}{4}\log\left|\dfrac{x-1}{x+1}\right|+\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$ (C) $\dfrac{1}{4}\log\left|\dfrac{x-1}{x+1}\right|-\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$ (D) $\dfrac{1}{4}\log\left|\dfrac{x+1}{x-1}\right|+\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$

   ›Reveal solutionSolution

   Splitting $\dfrac{x^2}{x^4-1}$ into a sum of $\dfrac{1}{x^2-1}$ and $\dfrac{1}{x^2+1}$ (halved) gives a standard log + arctan combination.

   Concept and Intuition

   Rather than doing full partial fractions with four unknowns, it's faster to notice $x^4-1=(x^2-1)(x^2+1)$ and that $\dfrac{1}{x^2-1}+\dfrac{1}{x^2+1}=\dfrac{(x^2+1)+(x^2-1)}{x^4-1}=\dfrac{2x^2}{x^4-1}$. This directly gives $\dfrac{x^2}{x^4-1}$ as half that sum — a shortcut avoiding solving for four separate constants.

   Step-by-Step Solution

   1. Write the denominator as $x^4-1=(x^2-1)(x^2+1)$.
   2. Observe: $\dfrac{1}{x^2-1}+\dfrac{1}{x^2+1}=\dfrac{2x^2}{x^4-1}$, so $\dfrac{x^2}{x^4-1}=\dfrac12\left[\dfrac{1}{x^2-1}+\dfrac{1}{x^2+1}\right]$.
   3. Use the standard integrals: $\displaystyle\int\dfrac{dx}{x^2-1}=\dfrac12\log\left|\dfrac{x-1}{x+1}\right|+c_1$ and $\displaystyle\int\dfrac{dx}{x^2+1}=\tan^{-1}x+c_2$.
   4. Combine: $\displaystyle\int\dfrac{x^2}{x^4-1}dx=\dfrac12\left[\dfrac12\log\left|\dfrac{x-1}{x+1}\right|+\tan^{-1}x\right]+c=\dfrac14\log\left|\dfrac{x-1}{x+1}\right|+\dfrac12\tan^{-1}x+c$.

   Common Mistakes …

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5. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\int\dfrac{2x^2-3}{(x^2-4)(x^2+1)}dx=A\tan^{-1}x+B\log(x-2)+C\log(x+2)$ then $6A+7B-5C=$ (A) $9$ (B) $10$ (C) $6$ (D) $8$

   ›Reveal solutionSolution

   Partial fractions of a rational function whose denominator has both real linear factors and an irreducible quadratic factor; the required integral form pins down the decomposition.

   Concept and Intuition

   The target antiderivative form $A\tan^{-1}x+B\log(x-2)+C\log(x+2)$ tells us exactly what partial-fraction decomposition must have produced it: a term $A/(x^2+1)$ (integrates to $A\tan^{-1}x$), and terms $B/(x-2)$, $C/(x+2)$.

   Step-by-Step Solution

   1. Write $\dfrac{2x^2-3}{(x-2)(x+2)(x^2+1)}=\dfrac{A}{x^2+1}+\dfrac{B}{x-2}+\dfrac{C}{x+2}$.
   2. Multiply through: $2x^2-3=A(x^2-4)+B(x+2)(x^2+1)+C(x-2)(x^2+1)$.
   3. Set $x=2$: $5=A(0)+B(4)(5)+0\Rightarrow 20B=5\Rightarrow B=\tfrac14$.
   4. Set $x=-2$: $5=0+0+C(-4)(5)\Rightarrow -20C=5\Rightarrow C=-\tfrac14$. …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The partial fraction decomposition of $\dfrac{x^4 + 24x^2 + 28}{(x^2+1)^3}$ is (A) $\dfrac{1}{x^2+1} - \dfrac{22}{(x^2+1)^2} + \dfrac{5}{(x^2+1)^3}$ (B) $\dfrac{1}{x^2+1} + \dfrac{22}{(x^2+1)^2} + \dfrac{5}{(x^2+1)^3}$ (C) $\dfrac{1}{x^2+1} - \dfrac{22}{(x^2+1)^2} - \dfrac{5}{(x^2+1)^3}$ (D) $\dfrac{1}{x^2+1} + \dfrac{22}{(x^2+1)^2} - \dfrac{5}{(x^2+1)^3}$

   ›Reveal solutionSolution

   This is a partial-fraction problem made easy by substituting $t=x^2+1$ so the whole numerator becomes a polynomial in $t$. Answer: all three signs are $+$, with coefficients $1,22,5$.

   Concept and Intuition

   Since the denominator is a power of $(x^2+1)$ only, and the numerator is a polynomial purely in $x^2$, it's far simpler to substitute $t=x^2+1$ (so $x^2=t-1$) and rewrite the numerator as a polynomial in $t$, rather than solving for unknown constants $A,B,C$ by matching coefficients directly in $x$.

   Step-by-Step Solution

   1. Let $t = x^2+1$, so $x^2 = t-1$, and $x^4 = (x^2)^2 = (t-1)^2 = t^2-2t+1$.
   2. Substitute into the numerator: $x^4+24x^2+28 = (t^2-2t+1) + 24(t-1) + 28$.
   3. Expand: $t^2-2t+1+24t-24+28 = t^2 + 22t + 5$.
   4. So the expression becomes $\dfrac{t^2+22t+5}{t^3} = \dfrac{t^2}{t^3}+\dfrac{22t}{t^3}+\dfrac{5}{t^3} = \dfrac{1}{t}+\dfrac{22}{t^2}+\dfrac{5}{t^3}$. …

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7. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.Given $\dfrac{3x-2}{(x+1)^2(x+3)} = \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2} + \dfrac{C}{x+3}$ then $4A + 2B + 4C$ (A) $5$ (B) $-5$ (C) $-3$ (D) $3$

   ›Reveal solutionSolution

   Standard partial-fraction cover-up plus a coefficient match pins down $A,B,C$, giving $4A+2B+4C=-5$.

   Concept and Intuition

   For a repeated linear factor $(x+1)^2$ together with a simple factor $(x+3)$, the cover-up (Heaviside) method quickly gives $B$ and $C$ by substituting the roots that make each factor vanish. The remaining constant $A$ is then found by matching a coefficient (here, the $x^2$ coefficient, since the numerator on the left has no $x^2$ term).

   Step-by-Step Solution

   1. Clear denominators: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$.
   2. Set $x=-1$ (kills the $A$ and $C$ terms): $3(-1)-2 = B(-1+3) \Rightarrow -5 = 2B \Rightarrow B=-\dfrac{5}{2}$.
   3. Set $x=-3$ (kills the $A$ and $B$ terms): $3(-3)-2 = C(-3+1)^2 \Rightarrow -11 = 4C \Rightarrow C=-\dfrac{11}{4}$.
   4. Match the coefficient of $x^2$ on both sides (LHS has none): $0 = A + C \Rightarrow A = \dfrac{11}{4}$. …

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8. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $\dfrac{x+2}{x^2-3}$ is one of the partial fractions of $\dfrac{3x^3 - x^2 - 2x + 17}{x^4 + x^2 - 12}$, then the other partial fraction of it is (A) $\dfrac{2x+3}{x^2-4}$ (B) $\dfrac{3x+2}{x^2+4}$ (C) $\dfrac{2x-3}{x^2+4}$ (D) $\dfrac{3x-2}{x^2-4}$

   ›Reveal solutionSolution

   This is a partial-fraction decomposition where one fraction is given; the other, over the irreducible quadratic $x^2+4$, is found by matching coefficients — the answer is $\dfrac{2x-3}{x^2+4}$.

   Concept and Intuition

   The quartic denominator factors as a product of two irreducible (over the reals, no rational roots) quadratics, $x^2-3$ and $x^2+4$. Since $x^2+4$ has no real roots, its partial fraction numerator must be a general linear expression $Ax+B$, not a constant. With one of the two fractions already known, the other is recovered by clearing denominators and matching the coefficients of the resulting polynomial identity.

   Step-by-Step Solution

   1. Factor the denominator: $x^4+x^2-12$. Let $u=x^2$: $u^2+u-12=(u+4)(u-3)$, so $x^4+x^2-12=(x^2+4)(x^2-3)$.
   2. Write $\dfrac{3x^3-x^2-2x+17}{(x^2-3)(x^2+4)}=\dfrac{x+2}{x^2-3}+\dfrac{Ax+B}{x^2+4}$.
   3. Multiply both sides by $(x^2-3)(x^2+4)$: $(x+2)(x^2+4)+(Ax+B)(x^2-3)=3x^3-x^2-2x+17$.
   4. Expand: $(x+2)(x^2+4)=x^3+2x^2+4x+8$, and $(Ax+B)(x^2-3)=Ax^3+Bx^2-3Ax-3B$. …

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9. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $\dfrac{2x^4 - x^3 + 3x^2 - x + 4}{x^2 - 3x + 2} = f(x) + \dfrac{A}{x-1} + \dfrac{B}{x-2}$ then ______ (A) $f(x) = 2x^2 + 5x + 14, A + B = 39$ (B) $f(x) = 2x^2 - 5x + 14, A + B = 31$ (C) $f(x) = 2x^2 + 5x + 14, A + B = 31$ (D) $f(x) = 2x^2 + 5x + 14, A = 4, B = 35$

   ›Reveal solutionSolution

   Long-divide the quartic by the quadratic to get $f(x)=2x^2+5x+14$ with remainder $31x-24$, then

   resolve the remainder into partial fractions to get $A=-7,\ B=38$, so $A+B=31$.

   Concept and Intuition

   Since the numerator's degree (4) exceeds the denominator's degree (2), the fraction splits into a

   polynomial part $f(x)$ (via polynomial long division) plus a proper-fraction remainder, which is

   then decomposed into partial fractions over the denominator's linear factors $(x-1)(x-2)$.

   Step-by-Step Solution

   1. Divide $2x^4-x^3+3x^2-x+4$ by $x^2-3x+2$:
      • $2x^4\div x^2=2x^2$; $2x^2(x^2-3x+2)=2x^4-6x^3+4x^2$; subtract: $5x^3-x^2-x+4$.
      • $5x^3\div x^2=5x$; $5x(x^2-3x+2)=5x^3-15x^2+10x$; subtract: $14x^2-11x+4$.
      • $14x^2\div x^2=14$; $14(x^2-3x+2)=14x^2-42x+28$; subtract: $31x-24$.
   2. So $f(x)=2x^2+5x+14$ and the remainder is $31x-24$ over $x^2-3x+2=(x-1)(x-2)$.
   3. Write $\dfrac{31x-24}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}$, so …

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10. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
    Q.$\frac{x^4}{(x^2+1)(x^2+3)} =$ (A) $\frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$ for some $A,B,C,D \in \mathbb{R}\setminus\{0\}$ (B) $\frac{Ax+B}{x^2+1} + \frac{Cx}{x^2+1}$ for some $A,B,C \in \mathbb{R}\setminus\{0\}$ (C) $\frac{Ax}{x^2+1} + \frac{Bx}{x^2+3}$ for some $A,B \in \mathbb{R}\setminus\{0\}$ (D) $1 + \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$ for some $A,B,C,D \in \mathbb{R}$

    ›Reveal solutionSolution

    Since numerator and denominator have equal degree (4 each), an extra constant "+1" term is required

    before the two proper partial fractions — matching option (D).

    Concept and Intuition

    Partial fraction decomposition applies directly only to a proper rational function (numerator

    degree strictly less than denominator degree). Here $(x^2+1)(x^2+3)$ expands to a degree-4 polynomial,

    exactly matching the numerator's degree 4 — so the fraction is improper, and we must first extract

    a polynomial part (here just a constant, since both are degree 4) via division, leaving a genuinely

    proper remainder to split over the two irreducible quadratic factors.

    Step-by-Step Solution

    1. Expand the denominator: $(x^2+1)(x^2+3) = x^4+4x^2+3$.
    2. Since numerator degree (4) = denominator degree (4), divide: $x^4 = 1\cdot(x^4+4x^2+3) - (4x^2+3)$.
    3. So $\dfrac{x^4}{(x^2+1)(x^2+3)} = 1 - \dfrac{4x^2+3}{(x^2+1)(x^2+3)}$.
    4. The remaining fraction $\dfrac{4x^2+3}{(x^2+1)(x^2+3)}$ is now proper and splits over the two distinct irreducible quadratics as $\dfrac{A'x+B'}{x^2+1}+\dfrac{C'x+D'}{x^2+3}$.
    5. Absorbing signs into new constants gives exactly the form $1+\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+3}$, …

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11. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.If $\dfrac{4x}{(x^2-1)^2} = \dfrac{A_1}{x-1} + \dfrac{A_2}{(x-1)^2} + \dfrac{A_3}{x+1} + \dfrac{A_4}{(x+1)^2}$, then $A_1 + A_2 + A_3 + A_4 =$ (A) $-2$ (B) $1$ (C) $0$ (D) $\dfrac{3}{2}$

    ›Reveal solutionSolution

    Clearing denominators and substituting convenient values of $x$ (the repeated roots plus two extra points) pins down all four constants; they add up to $0$.

    Concept and Intuition

    For a partial fraction decomposition with repeated linear factors, substituting the roots directly isolates the "squared-term" coefficients instantly, while substituting a couple of extra convenient values (like $x=0$ and $x=2$) gives enough equations to solve for the remaining linear-term coefficients.

    Step-by-Step Solution

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

    $4x=A_1(x-1)(x+1)^2+A_2(x+1)^2+A_3(x+1)(x-1)^2+A_4(x-1)^2.$

    2. Set $x=1$: $4=A_2(2)^2=4A_2 \Rightarrow A_2=1$.
    3. Set $x=-1$: $-4=A_4(-2)^2=4A_4 \Rightarrow A_4=-1$.
    4. Set $x=0$: $0=A_1(-1)(1)+A_2(1)+A_3(1)(1)+A_4(1)=-A_1+A_2+A_3+A_4$. Using $A_2=1,A_4=-1$: $0=-A_1+A_3 \Rightarrow A_1=A_3$. …

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12. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.If $\dfrac{3x+1}{(x-1)(x^2+2)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+2}$, then $5(A-B)=$ (A) $A+C$ (B) $8C$ (C) $C+8$ (D) $\dfrac{C}{8}$

    ›Reveal solutionSolution

    This tests partial-fraction decomposition and coefficient comparison. Answer: $5(A-B)=8C$.

    Concept and Intuition

    To find $A,B,C$ in a partial fraction decomposition, clear denominators to get a polynomial identity, then either substitute convenient values of $x$ (like the root of the linear factor) or compare coefficients of like powers of $x$.

    Step-by-Step Solution

    1. Clear denominators: $3x+1 = A(x^2+2)+(Bx+C)(x-1)$.
    2. Substitute $x=1$ (kills the $(Bx+C)(x-1)$ term): $3(1)+1 = A(1+2) \Rightarrow 4=3A \Rightarrow A=\dfrac43$.
    3. Expand the right side: $Ax^2+2A+Bx^2-Bx+Cx-C = (A+B)x^2+(C-B)x+(2A-C)$.
    4. Compare coefficient of $x^2$ (LHS has none): $A+B=0\Rightarrow B=-A=-\dfrac43$.
    5. Compare coefficient of $x^1$: $C-B=3\Rightarrow C=3+B=3-\dfrac43=\dfrac53$.
    6. Check constant term: $2A-C = \dfrac83-\dfrac53=1$ ✓, consistent. …

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