Q.Evaluate: $\int \dfrac{x}{x^4-1}\,dx$

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 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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3. 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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4. 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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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-2026-05-12-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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7. 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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8. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If $\dfrac{1}{x^4+1} = \dfrac{Ax+B}{x^2+\sqrt{2}x+1} + \dfrac{Cx+D}{x^2-\sqrt{2}x+1}$ then $BD - AC =$ (A) $\dfrac{3}{8}$ (B) $\dfrac{1}{8}$ (C) 1 (D) 0

   ›Reveal solutionSolution

   This tests partial-fraction decomposition of $\dfrac{1}{x^4+1}$ over its two real quadratic factors; the answer is $\dfrac38$.

   Concept and Intuition

   $x^4+1$ factors as a difference of squares: $(x^2+1)^2-(\sqrt2 x)^2 = (x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$. Matching coefficients after clearing denominators pins down $A,B,C,D$ uniquely.

   Step-by-Step Solution

   1. Write $1=(Ax+B)(x^2-\sqrt2x+1)+(Cx+D)(x^2+\sqrt2x+1)$.
   2. Expand and collect by power of $x$:
      • $x^3$: $A+C=0 \Rightarrow C=-A$
      • $x^2$: $(B-\sqrt2A)+(D+\sqrt2C)=0$. With $C=-A$ this gives $B+D-2\sqrt2A=0$
      • $x^1$: $(A-\sqrt2B)+(C+\sqrt2D)=0$. With $C=-A$ this gives $\sqrt2(D-B)=0\Rightarrow D=B$
      • $x^0$: $B+D=1$
   3. From $D=B$ and $B+D=1$: $B=D=\dfrac12$.
   4. From $B+D=2\sqrt2A$: $1=2\sqrt2A \Rightarrow A=\dfrac{1}{2\sqrt2}=\dfrac{\sqrt2}{4}$, and $C=-A=-\dfrac{\sqrt2}{4}$. …

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9. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.If $\dfrac{x}{(x-1)(x^2+1)^2} = \dfrac{1}{4}\left[\dfrac{1}{x-1} - \dfrac{x+1}{x^2+1}\right]+y$, then $y =$ (A) $\dfrac{1}{2}\left[\dfrac{1-x}{(x^2+1)^2}\right]$ (B) $\dfrac{1+x}{3(x^2+1)^2}$ (C) $\dfrac{1-x}{(x^2-1)^2}$ (D) $\dfrac{1+x}{(x^2+1)^2}$

   ›Reveal solutionSolution

   This tests partial fraction decomposition with a repeated irreducible quadratic factor, then matching the remaining unaccounted term to $y$.

   Concept and Intuition

   For a denominator $(x-1)(x^2+1)^2$, the full decomposition has the form $\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}$. The problem already gives the first two pieces combined as $\tfrac14\left[\tfrac{1}{x-1}-\tfrac{x+1}{x^2+1}\right]$, so $y$ must be exactly the third piece.

   Step-by-Step Solution

   1. Write $\dfrac{x}{(x-1)(x^2+1)^2} = \dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}$.
   2. Multiply through: $x = A(x^2+1)^2 + (Bx+C)(x-1)(x^2+1) + (Dx+E)(x-1)$.
   3. Set $x=1$: $1 = 4A \Rightarrow A=\tfrac14$.
   4. Expand and match coefficients of $x^4,x^3,x^2,x^1,x^0$: this yields $B=-\tfrac14,\ C=-\tfrac14,\ D=-\tfrac12,\ E=\tfrac12$.
   5. So the full decomposition is $\dfrac{1/4}{x-1} - \dfrac14\cdot\dfrac{x+1}{x^2+1} + \dfrac{(1-x)/2}{(x^2+1)^2}$. …

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10. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.If $\dfrac{x}{(x^2+1)^2(x-1)} = \dfrac{Ax+B}{x^2+1} + \dfrac{Cx+D}{(x^2+1)^2} + \dfrac{E}{x-1}$, then $A + B - C + 2D =$ (A) $\dfrac{1}{2}$ (B) $1$ (C) $\dfrac{3}{2}$ (D) $2$

    ›Reveal solutionSolution

    This tests standard partial-fraction decomposition with a repeated irreducible quadratic factor, using a mix of "plug in a root" and "match coefficients" techniques. The final computed value is $A+B-C+2D=1$.

    Concept and Intuition

    When the denominator has an irreducible quadratic factor repeated twice, $(x^2+1)^2$, along with a simple linear factor $(x-1)$, the partial fraction form needs a linear numerator ($Ax+B$, $Cx+D$) over each power of the quadratic, plus a constant ($E$) over the linear factor. The cleanest way to solve is: clear denominators, plug in the linear factor's root to isolate $E$ instantly, then expand the rest and match coefficients of each power of $x$ to get the remaining unknowns.

    Step-by-Step Solution

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

    $x = (Ax+B)(x^2+1)(x-1) + (Cx+D)(x-1) + E(x^2+1)^2.$

    2. Find $E$ quickly: set $x=1$. The first two terms vanish (each has a factor of $(x-1)$), leaving $1 = E(1^2+1)^2 = 4E$, so $E=\dfrac14$.
    3. Expand the rest. First, $(x^2+1)(x-1) = x^3-x^2+x-1$, so

    $(Ax+B)(x^3-x^2+x-1) = Ax^4+(-A+B)x^3+(A-B)x^2+(-A+B)x-B.$

    Next, $(Cx+D)(x-1) = Cx^2+(D-C)x-D$. And $E(x^2+1)^2 = Ex^4+2Ex^2+E$.

    4. Collect coefficients by power of $x$ and equate to the right-hand side of the original equation (which is just $x$, so coefficients are $0,0,0,1,0$ for $x^4,x^3,x^2,x^1,x^0$ respectively):

    • $x^4$: $A+E=0 \Rightarrow A=-E=-\dfrac14$.
    • $x^3$: $-A+B=0 \Rightarrow B=A=-\dfrac14$.
    • $x^2$: $(A-B)+C+2E=0$. Since $A=B$, this gives $C=-2E=-\dfrac12$. …

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11. 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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12. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.Which of the following is an improper rational traction? (A) $\dfrac{x^2+1}{(x^2+2)(x^2+x+1)}$ (B) $\dfrac{x^2+1}{(x+3)(x^2-x+1)}$ (C) $\dfrac{x}{(x^2+3x+1)}$ (D) $\dfrac{x^2+1}{x^2-1}$

    ›Reveal solutionSolution

    A rational fraction is proper only when the numerator's degree is strictly less than the denominator's; option (D) has numerator and denominator of the same degree (2 and 2), making it the only improper fraction listed.

    Concept and Intuition

    In partial fraction decomposition, a "proper" rational fraction has numerator degree strictly less than denominator degree, ensuring the fraction tends to zero as $x\to\infty$ and decomposes cleanly into partial fractions without needing polynomial long division first. An "improper" fraction (numerator degree $\ge$ denominator degree) must first be reduced via division into a polynomial plus a proper remainder fraction.

    Step-by-Step Solution

    1. Option (A): $\dfrac{x^2+1}{(x^2+2)(x^2+x+1)}$ — numerator degree 2; denominator is a product of two quadratics, degree $2+2=4$. Since $2<4$, this is proper.
    2. Option (B): $\dfrac{x^2+1}{(x+3)(x^2-x+1)}$ — numerator degree 2; denominator is linear times quadratic, degree $1+2=3$. Since $2<3$, proper.
    3. Option (C): $\dfrac{x}{x^2+3x+1}$ — numerator degree 1; denominator degree 2. Since $1<2$, proper. …

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