Q.Find $\int \dfrac{x^2}{(x^2+1)(x^2+4)}\, dx$

Showing the 12 most recent of 63 on this concept.

1. 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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2. 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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3. 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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4. 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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5. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $\dfrac{x^2}{(x^2+2)(x^4-1)}=\dfrac{A}{x^2-1}+\dfrac{B}{x^2+1}+\dfrac{C}{x^2+2}$, then $A+B-C=$ (A) $0$ (B) $\dfrac{4}{3}$ (C) $\dfrac{3}{4}$ (D) $2$

   ›Reveal solutionSolution

   A partial-fractions problem in disguise (substitute $y=x^2$); solving gives $A+B-C=\frac{4}{3}$.

   Concept and Intuition

   Since $x^4-1=(x^2-1)(x^2+1)$, the whole expression is a rational function purely in $y=x^2$. Substituting $y=x^2$ converts it into an ordinary partial-fractions decomposition with three distinct linear factors $(y-1),(y+1),(y+2)$, solvable by the cover-up (Heaviside) method.

   Step-by-Step Solution

   1. Let $y=x^2$. The equation becomes $\dfrac{y}{(y+2)(y-1)(y+1)}=\dfrac{A}{y-1}+\dfrac{B}{y+1}+\dfrac{C}{y+2}$.
   2. Clear denominators: $y=A(y+1)(y+2)+B(y-1)(y+2)+C(y-1)(y+1)$.
   3. At $y=1$: $1=A(2)(3)=6A\Rightarrow A=\dfrac16$.
   4. At $y=-1$: $-1=B(-2)(1)=-2B\Rightarrow B=\dfrac12$.
   5. At $y=-2$: $-2=C(-3)(-1)=3C\Rightarrow C=-\dfrac23$. …

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6. 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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7. 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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8. 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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9. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.$\int \dfrac{x-1}{(x-2)(x-3)}\,dx=$ (A) $2\log|x-3|+\log|x-2|+c$ (B) $\log|x-3|-\log|x-2|+c$ (C) $\log|x-3|^2-\log|x+2|+c$ (D) $\log\left|\dfrac{(x-3)^2}{x-2}\right|+c$

   ›Reveal solutionSolution

   A rational function with distinct linear factors in the denominator — resolve into partial fractions, integrate each term as a log, then combine using log rules.

   Concept and Intuition

   Any proper rational function with distinct linear denominator factors can be split into simple fractions $\frac{A}{x-a}+\frac{B}{x-b}$, each of which integrates to $A\log|x-a|$. Combining the two logs at the end into a single log of a ratio/power lets you match against answer choices written as one combined logarithm.

   Step-by-Step Solution

   1. Write $\dfrac{x-1}{(x-2)(x-3)} = \dfrac{A}{x-2}+\dfrac{B}{x-3}$, so $x-1 = A(x-3)+B(x-2)$.
   2. Put $x=2$: $2-1 = A(2-3) \Rightarrow 1 = -A \Rightarrow A=-1$.
   3. Put $x=3$: $3-1 = B(3-2) \Rightarrow 2=B$.
   4. So $\displaystyle\int\frac{x-1}{(x-2)(x-3)}dx = -\log|x-2| + 2\log|x-3| + c$. …

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10. 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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11. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.$\int \frac{dx}{\sin x + \sin 2x} =$ (A) $\frac{1}{4}\log|1-\cos x| + \frac{1}{3}\log|1+\cos x| - \frac{2}{3}\log|1+\cos 2x| + c$ (B) $\frac{1}{3}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{1}{3}\log|1+\cos 2x| + c$ (C) $\frac{1}{6}\log|1-\cos x| + \frac{1}{2}\log|1+\cos x| - \frac{2}{3}\log|1+2\cos x| + c$ (D) $\frac{1}{6}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{2}{3}\log|1+2\cos x| + c$

    ›Reveal solutionSolution

    Factoring $\sin x+\sin2x=\sin x(1+2\cos x)$ and substituting $t=\cos x$ reduces this to a rational-function partial-fractions integral, giving $\frac16\log|1-\cos x|+\frac12\log|1+\cos x|-\frac23\log|1+2\cos x|+c$.

    Concept and Intuition

    Whenever an integral has $\sin x$ (an odd power effectively) times other cosine factors in the denominator, multiplying numerator and denominator by $\sin x$ turns $\sin^2x$ into $1-\cos^2x$, which lets us substitute $t=\cos x$ and reduce the whole problem to partial fractions of a rational function in $t$ — a completely mechanical final step.

    Step-by-Step Solution

    1. Factor the denominator: $\sin x+\sin2x = \sin x + 2\sin x\cos x = \sin x(1+2\cos x)$.
    2. So the integral is $\displaystyle\int\frac{dx}{\sin x(1+2\cos x)}$.
    3. Multiply top and bottom by $\sin x$: $\displaystyle\int\frac{\sin x\,dx}{\sin^2x(1+2\cos x)}=\int\frac{\sin x\,dx}{(1-\cos^2x)(1+2\cos x)}=\int\frac{\sin x\,dx}{(1-\cos x)(1+\cos x)(1+2\cos x)}$.
    4. Substitute $t=\cos x$, $dt=-\sin x\,dx$: integral $=-\displaystyle\int\frac{dt}{(1-t)(1+t)(1+2t)}$.
    5. Partial fractions: $\dfrac{1}{(1-t)(1+t)(1+2t)}=\dfrac{A}{1-t}+\dfrac{B}{1+t}+\dfrac{C}{1+2t}$. Evaluating at $t=1$: $A=\tfrac16$. At $t=-1$: $B=\tfrac12$. At $t=-\tfrac12$: $C=-\tfrac23$. …

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12. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{x+1}{x^3-1}\,dx=$ (A) $\dfrac{1}{3}\log\left(\dfrac{x+1}{x^2+x+1}\right)+c$ (B) $\dfrac{1}{3}\log\left(\dfrac{(x-1)^2}{x^2+x+1}\right)+c$ (C) $\dfrac{1}{3}\log\left(\dfrac{x-1}{x^2+x+1}\right)+c$ (D) $\dfrac{1}{3}\log\left(\dfrac{(x+1)^2}{x^2-x+1}\right)+c$

    ›Reveal solutionSolution

    This is a rational-function integral solved by partial fractions after factoring $x^3-1$; the result combines into a single log of $\dfrac{(x-1)^2}{x^2+x+1}$.

    Concept and Intuition

    Factor the cubic denominator as difference of cubes, then split into a linear-factor term (giving a plain log) and an irreducible-quadratic term (giving a log plus, here, no arctangent term since the numerator works out to be an exact multiple of the quadratic's derivative).

    Step-by-Step Solution

    1. $x^3-1=(x-1)(x^2+x+1)$.
    2. Write $\dfrac{x+1}{(x-1)(x^2+x+1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+x+1}$.
    3. $x+1=A(x^2+x+1)+(Bx+C)(x-1)$. At $x=1$: $2=3A\Rightarrow A=\frac23$.
    4. Matching $x^2$: $A+B=0\Rightarrow B=-\frac23$. Matching constants: $A-C=1\Rightarrow C=-\frac13$.
    5. $\displaystyle\int\frac{2/3}{x-1}dx=\frac23\log|x-1|$.
    6. $\displaystyle\int\frac{-\frac23x-\frac13}{x^2+x+1}dx=-\frac13\int\frac{2x+1}{x^2+x+1}dx=-\frac13\log(x^2+x+1)$ (numerator is exactly the derivative of the denominator). …

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