Q.Integrate the following function: $\frac{2x}{(x^2+1)(x^2+3)}$

Showing the 12 most recent of 63 on this concept.

1. 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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2. 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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3. 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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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 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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6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.One of the partial fractions of $\dfrac{2x^2+x-3}{(x^2+2)(3x-1)}$ is (A) $\dfrac{22}{19(3x-1)}$ (B) $\dfrac{20x-13}{19(x^2+2)}$ (C) $\dfrac{20x+13}{19(x^2+2)}$ (D) $\dfrac{22}{3x-1}$

   ›Reveal solutionSolution

   Standard partial-fraction decomposition with one irreducible quadratic factor and one linear factor. Solving for the constants gives $A=\frac{20}{19}, B=\frac{13}{19}, C=-\frac{22}{19}$, so the quadratic-denominator fraction is $\frac{20x+13}{19(x^2+2)}$.

   Concept and Intuition

   Since $x^2+2$ has no real roots, it contributes a fraction with a linear numerator $Ax+B$, while the linear factor $3x-1$ contributes a constant numerator $C$. Clearing denominators turns the problem into matching coefficients (or, more efficiently, substituting convenient values of $x$ — especially the root of the linear factor, which instantly isolates $C$).

   Step-by-Step Solution

   1. Set up: $2x^2+x-3 = (Ax+B)(3x-1) + C(x^2+2)$.
   2. Substitute $x=\tfrac13$ (root of $3x-1$), which kills the $(Ax+B)(3x-1)$ term:

   $2\left(\tfrac19\right)+\tfrac13-3 = C\left(\tfrac19+2\right) \Rightarrow \frac{2+3-27}{9}=C\cdot\frac{19}{9}\Rightarrow -\frac{22}{9}=\frac{19C}{9}\Rightarrow C=-\frac{22}{19}.$

   3. Substitute $x=0$: LHS $=-3$; RHS $=B(-1)+2C = -B+2\left(-\tfrac{22}{19}\right)=-B-\tfrac{44}{19}$.

   $-3=-B-\frac{44}{19}\Rightarrow B=\frac{44}{19}-3=\frac{44-57}{19}=-\frac{13}{19}...$

   Recheck sign: $-3 = -B - \frac{44}{19} \Rightarrow -B = -3+\frac{44}{19}=\frac{-57+44}{19}=-\frac{13}{19} \Rightarrow B=\frac{13}{19}.$

   4. Substitute $x=1$: LHS $=2+1-3=0$; RHS $=(A+B)(2)+3C = 2A+2B+3C$.

   $0 = 2A + 2\left(\frac{13}{19}\right)+3\left(-\frac{22}{19}\right) = 2A+\frac{26-66}{19}=2A-\frac{40}{19}\Rightarrow A=\frac{20}{19}.$ …

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7. 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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8. 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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9. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $\dfrac{x^3}{(2x-1)(x+2)(x-3)} = A + \dfrac{B}{2x-1} + \dfrac{C}{x+2} + \dfrac{D}{x-3}$ then $A =$ (A) $\dfrac{1}{2}$ (B) $\dfrac{-1}{50}$ (C) $\dfrac{-8}{25}$ (D) $\dfrac{27}{25}$

   ›Reveal solutionSolution

   As $x\to\infty$, the left side tends to $1/2$ (ratio of leading coefficients) while the right side tends to $A$; hence $A=1/2$.

   Concept and Intuition

   Here the numerator's degree (3) equals the denominator's degree (3), so a genuine partial-fraction decomposition needs a polynomial (constant, here) term $A$ in addition to the proper-fraction terms. The value of $A$ can be found quickly by comparing leading behaviour as $x\to\infty$.

   Step-by-Step Solution

   1. Expand the denominator: $(2x-1)(x+2)(x-3) = 2x^3-3x^2-11x+6$.
   2. As $x\to\infty$: $\dfrac{x^3}{2x^3-3x^2-11x+6}\to\dfrac{1}{2}$.
   3. On the right side, as $x\to\infty$, all the proper-fraction terms $\dfrac{B}{2x-1},\dfrac{C}{x+2},\dfrac{D}{x-3}\to0$, leaving just $A$. …

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10. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.$\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \Rightarrow 7A+2B+C =$ (A) $8$ (B) $9$ (C) $10$ (D) $11$

    ›Reveal solutionSolution

    Clear denominators in the partial fraction decomposition, use $x=1$ to find $A$ quickly, then match coefficients for $B$ and $C$.

    Concept and Intuition

    Since $x^3-1=(x-1)(x^2+x+1)$, the partial fraction form given is standard. Plugging in the root of the linear factor ($x=1$) isolates $A$ immediately; matching remaining coefficients (or plugging in more values) gives $B,C$.

    Step-by-Step Solution

    1. Multiply both sides by $x^3-1=(x-1)(x^2+x+1)$: $2x^2+1 = A(x^2+x+1) + (Bx+C)(x-1)$.
    2. Set $x=1$: $2(1)+1 = A(1+1+1) + 0 \Rightarrow 3 = 3A \Rightarrow A=1$.
    3. Expand the right side: $A x^2+Ax+A + Bx^2-Bx+Cx-C = (A+B)x^2+(A-B+C)x+(A-C)$.
    4. Match $x^2$ coefficient: $A+B=2 \Rightarrow B = 2-1=1$.
    5. Match constant term: $A - C = 1 \Rightarrow C = A-1 = 0$. …

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11. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.$\dfrac{x^2+1}{x^4+4} = \dfrac{Ax+B}{x^2-2x+2} + \dfrac{Cx+D}{x^2+2x+2} \Rightarrow 3A+2B+3C =$ (A) -D (B) D (C) 2D (D) -2D

    ›Reveal solutionSolution

    Recognizing the Sophie Germain factorization of $x^4+4$ turns this into a routine partial-fractions coefficient match.

    Concept and Intuition

    $x^4+4a^4=(x^2-2ax+2a^2)(x^2+2ax+2a^2)$ is the Sophie Germain identity; with $a=1$ it gives exactly the two quadratic factors seen in the denominators here.

    Step-by-Step Solution

    1. Confirm $x^4+4=(x^2-2x+2)(x^2+2x+2)$ — the given denominators are exactly this factorization.
    2. Write $x^2+1=(Ax+B)(x^2+2x+2)+(Cx+D)(x^2-2x+2)$.
    3. Expand: $(Ax+B)(x^2+2x+2)=Ax^3+(2A+B)x^2+(2A+2B)x+2B$.
    4. $(Cx+D)(x^2-2x+2)=Cx^3+(D-2C)x^2+(2C-2D)x+2D$.
    5. Sum and match to $0x^3+1x^2+0x+1$:
       • $x^3$: $A+C=0\Rightarrow C=-A$.
       • $x^0$: $2B+2D=1$.
       • $x^1$: $2A+2B+2C-2D=0\Rightarrow$ (using $C=-A$) $2B-2D=0\Rightarrow B=D$.
       • Then from $2B+2D=1$ with $B=D$: $4B=1\Rightarrow B=D=\tfrac14$. …

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12. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.If the equivalent partial fraction of $\frac{x^3}{(2x-1)(x+2)(x-3)}$ is of the form $A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$ then the value of $A+B+C =$ (A) $-8/25$ (B) $4/25$ (C) $-1/50$ (D) $1/2$

    ›Reveal solutionSolution

    This is an improper partial fraction (numerator degree = denominator degree), so there's a constant term $A$ found from the leading behaviour, and $B,C$ are found by the standard cover-up (root-substitution) method — giving $A+B+C=4/25$.

    Concept and Intuition

    When the degree of the numerator equals the degree of the denominator, ordinary partial fractions leave a nonzero polynomial part (here, just a constant $A$, since both degrees are 3 and the denominator's leading coefficient is 2 — so $A$ equals the ratio of leading coefficients, $1/2$). The remaining proper-fraction coefficients ($B$, $C$, $D$) are then found efficiently using the "cover-up" trick: multiply through by the denominator and substitute each root of a linear factor to instantly isolate that factor's coefficient.

    Step-by-Step Solution

    1. As $x\to\infty$, $\dfrac{x^3}{(2x-1)(x+2)(x-3)} \to \dfrac{x^3}{2x^3} = \dfrac12$, so the constant part is $A=\dfrac12$.
    2. Multiply both sides by $(2x-1)(x+2)(x-3)$: $x^3 = A(2x-1)(x+2)(x-3) + B(x+2)(x-3) + C(2x-1)(x-3) + D(2x-1)(x+2)$.
    3. Set $x=\tfrac12$ (kills the $A$, $C$, $D$ terms): $\left(\tfrac12\right)^3 = B\left(\tfrac12+2\right)\left(\tfrac12-3\right) = B(2.5)(-2.5)=-6.25B$. $\tfrac18 = -\tfrac{25}{4}B \Rightarrow B = -\tfrac{1}{50}$. …

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