Q.$\int \dfrac{x^2\,dx}{(x^2+a^2)(x^2+b^2)}$

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 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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3. 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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4. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If $\dfrac{A}{x-a} + \dfrac{Bx+C}{x^2+b^2} = \dfrac{1}{(x-a)(x^2+b^2)}$ then $C =$ (A) $\dfrac{-1}{a^2+b^2}$ (B) $\dfrac{1}{a^2+b^2}$ (C) $\dfrac{-a}{a^2+b^2}$ (D) $\dfrac{a}{a^2+b^2}$

   ›Reveal solutionSolution

   A standard partial-fractions comparison; clearing denominators and matching coefficients gives $C=\dfrac{-a}{a^2+b^2}$.

   Concept and Intuition

   When a rational function is split into partial fractions, multiplying through by the common denominator turns the identity into a polynomial identity, which must hold for all $x$. That lets us either substitute convenient values of $x$ or match coefficients of like powers of $x$.

   Step-by-Step Solution

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

   $A(x^2+b^2) + (Bx+C)(x-a) = 1$

   2. Put $x=a$ (kills the second term): $A(a^2+b^2)=1\Rightarrow A=\dfrac{1}{a^2+b^2}$.
   3. Expand the right side: $Ax^2+Ab^2+Bx^2-aBx+Cx-aC=1$.
   4. Coefficient of $x^2$: $A+B=0\Rightarrow B=-A=\dfrac{-1}{a^2+b^2}$. …

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5. 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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6. 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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7. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $\dfrac{17x-2}{12x^2-x-20} = \dfrac{A}{ax+5} + \dfrac{B}{3x+b}$ then $a.A + b.B =$ (A) 0 (B) 4 (C) 7 (D) 10

   ›Reveal solutionSolution

   Factoring the denominator and doing a partial-fraction decomposition gives $a=4,A=3,b=-4,B=2$, so $aA+bB=12-8=4$.

   Concept and Intuition

   Partial fraction decomposition requires first factoring the denominator to match the given linear-factor forms, then solving for the numerator constants by substituting the roots of each factor (the Heaviside cover-up method).

   Step-by-Step Solution

   1. Factor $12x^2-x-20$: looking for factors of $(4x+5)(3x-4)=12x^2-16x+15x-20=12x^2-x-20$. ✓
   2. Matching to the given form $\dfrac{A}{ax+5}+\dfrac{B}{3x+b}$: since one factor is $(4x+5)$, we get $a=4$; the other factor is $(3x-4)$, matching $3x+b$ gives $b=-4$.
   3. So $\dfrac{17x-2}{(4x+5)(3x-4)}=\dfrac{A}{4x+5}+\dfrac{B}{3x-4}$, i.e. $17x-2=A(3x-4)+B(4x+5)$.
   4. Set $x=\dfrac43$ (root of $3x-4=0$): $17\cdot\dfrac43-2=B\left(4\cdot\dfrac43+5\right)\Rightarrow\dfrac{68}{3}-\dfrac63=B\cdot\dfrac{31}{3}\Rightarrow\dfrac{62}{3}=\dfrac{31B}{3}\Rightarrow B=2$. …

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8. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\dfrac{x^2-7x+2}{x^4+3x^2+4} = \dfrac{Ax+B}{x^2+ax+2} + \dfrac{Cx+D}{x^2+bx+2}$ and $a>b$, then $B+D=$ (A) $a+b$ (B) $2a+b$ (C) $a+2b$ (D) $a-b$

   ›Reveal solutionSolution

   The quartic factors as a product of two quadratics with $a=1,b=-1$; comparing coefficients in the partial-fraction identity gives $B+D=1$, which equals $2a+b$.

   Concept and Intuition

   $x^4+3x^2+4$ has no real roots but factors nicely as a "sum/difference" trick: $x^4+3x^2+4=(x^2+2)^2-x^2=(x^2+x+2)(x^2-x+2)$, exactly matching the form $x^2+ax+2$ and $x^2+bx+2$ used in the partial fraction split.

   Step-by-Step Solution

   1. Write $x^4+3x^2+4=(x^2+2)^2-x^2=(x^2+2-x)(x^2+2+x)=(x^2-x+2)(x^2+x+2)$.
   2. Comparing with $(x^2+ax+2)(x^2+bx+2)$ and using $a>b$: $a=1,\ b=-1$.
   3. Expand $(Ax+B)(x^2-x+2)+(Cx+D)(x^2+x+2)$ and collect coefficients of $x^3,x^2,x,1$:
      • $x^3$: $A+C=0$
      • $x^2$: $(B-A)+(C+D)=1$
      • $x^1$: $(2A-B)+(2C+D)=-7$
      • $x^0$: $2B+2D=2$ …

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9. 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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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 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If $\dfrac{13x+43}{2x^2+17x+30} = \dfrac{A}{2x+5} + \dfrac{B}{x+6}$, then $A^2+B^2 =$ (A) $22/3$ (B) $52$ (C) $34$ (D) $18/5$

    ›Reveal solutionSolution

    Factoring the denominator and matching coefficients gives $A=3$, $B=5$, so $A^2+B^2=34$.

    Concept and Intuition

    Partial fraction decomposition of a rational function with a factorable denominator reduces to writing the numerator as a linear combination of the "cleared" denominators of each partial fraction, then matching coefficients of like powers of $x$ (or substituting convenient values of $x$).

    Step-by-Step Solution

    1. Factor $2x^2+17x+30$: looking for factors of the form $(2x+5)(x+6)$, expand to check: $(2x+5)(x+6) = 2x^2+12x+5x+30 = 2x^2+17x+30$. ✓.
    2. Write $\dfrac{13x+43}{(2x+5)(x+6)} = \dfrac{A}{2x+5}+\dfrac{B}{x+6}$. Clearing denominators: $13x+43 = A(x+6) + B(2x+5)$.
    3. Expand the right side: $Ax+6A+2Bx+5B = (A+2B)x + (6A+5B)$.
    4. Match coefficients: $A+2B=13$ (coefficient of $x$) and $6A+5B=43$ (constant term). …

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12. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If $\dfrac{ax+5}{(x^2+b)(x+3)} = \dfrac{x+21}{12(x^2+b)} + \dfrac{c}{12(x+3)}$, then $b^2 =$ (A) $a^3-c$ (B) $a^2+c$ (C) $a-c$ (D) $a+c$

    ›Reveal solutionSolution

    Clear denominators in the partial-fraction identity, match coefficients of powers of $x$ to solve for $a$, $b$, $c$, then check which combination of $a,c$ equals $b^2$. Answer: $b^2=a^3-c$.

    Concept and Intuition

    A partial-fraction identity must hold for all $x$, so after clearing denominators the two sides are polynomials that must agree coefficient-by-coefficient. This converts a rational-function identity into a simple linear system.

    Step-by-Step Solution

    1. Multiply both sides by $12(x^2+b)(x+3)$:

    $12(ax+5) = (x+21)(x+3) + c(x^2+b)$

    2. Expand the right side: $(x+21)(x+3)=x^2+24x+63$, so RHS $=x^2+24x+63+cx^2+cb=(1+c)x^2+24x+(63+cb)$.
    3. LHS $=12ax+60$ has no $x^2$ term, so matching $x^2$ coefficients: $1+c=0\Rightarrow c=-1$.
    4. Matching $x$ coefficients: $12a=24\Rightarrow a=2$.
    5. Matching constants: $60=63+cb=63+(-1)b=63-b\Rightarrow b=3$. …

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