Q.Integrate the following function: $\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$

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 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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3. 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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4. 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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5. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $\dfrac{x+2}{(x^2+3)(x^4+x^2)(x^2+2)} = \dfrac{Ax+B}{x^2+3} + \dfrac{Cx+D}{x^2+2} + \dfrac{Ex^3+Fx^2+Gx+H}{x^4+x^2}$ then $(E+F)(C+D)(A) =$ (A) $-\dfrac{1}{4}$ (B) $-\dfrac{3}{4}$ (C) $\dfrac{3}{4}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Cover-up on each factor gives $A=-\tfrac16$, $C=\tfrac12,\,D=1$, $E=-\tfrac13,\,F=-\tfrac23$; hence $(E+F)(C+D)(A)=\tfrac14$.

   The denominator factors as

   $(x^2+3)(x^4+x^2)(x^2+2),\qquad x^4+x^2=x^2(x^2+1).$

   Coefficient $A$ (from $x^2+3$). Multiply by $(x^2+3)$ and put $x^2=-3$:

   $Ax+B=\frac{x+2}{(x^4+x^2)(x^2+2)}\Big|_{x^2=-3}=\frac{x+2}{(6)(-1)}=-\frac{x+2}{6},$

   since $x^4+x^2=(-3)(-2)=6$ and $x^2+2=-1$. Thus $A=-\tfrac16$.

   Coefficients $C,D$ (from $x^2+2$). Multiply by $(x^2+2)$ and put $x^2=-2$:

   $Cx+D=\frac{x+2}{(x^4+x^2)(x^2+3)}\Big|_{x^2=-2}=\frac{x+2}{(2)(1)}=\frac{x}{2}+1,$

   since $x^4+x^2=(-2)(-1)=2$ and $x^2+3=1$. Thus $C=\tfrac12,\ D=1$, so $C+D=\tfrac32$.

   Coefficients $E,F$ (from $x^4+x^2=x^2(x^2+1)$). Multiply by $(x^4+x^2)$; at its roots the other terms vanish, so

   $Ex^3+Fx^2+Gx+H=\frac{x+2}{(x^2+3)(x^2+2)}\quad\text{at }x=0\text{ (double) and }x=\pm i.$ …

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6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\dfrac{x^4}{(x-1)(x-2)}=f(x)+\dfrac{A}{x-1}+\dfrac{B}{x-2}$, then $f(-2)+A+B=$ (A) 32 (B) 28 (C) 22 (D) 20

   ›Reveal solutionSolution

   This tests polynomial long division combined with partial fractions — split $\dfrac{x^4}{(x-1)(x-2)}$ into a polynomial part $f(x)$ plus proper fractions. Answer: $f(-2)+A+B=20$.

   Concept and Intuition

   When the numerator's degree ($4$) is greater than or equal to the denominator's degree ($2$), a rational function isn't purely a sum of partial fractions — you first must divide out a polynomial quotient $f(x)$, leaving a proper-fraction remainder that partial-fractions cleanly. Here $f(x)$ is exactly that quotient (degree $4-2=2$).

   Step-by-Step Solution

   1. Divide $x^4$ by $x^2-3x+2$ (long division): $x^4 = (x^2-3x+2)(x^2+3x+7) + (15x-14)$ Check: $(x^2-3x+2)(x^2+3x+7) = x^4 - 15x+14$, so adding $15x-14$ recovers $x^4$. ✓
   2. So $f(x)=x^2+3x+7$, and the remainder gives $\dfrac{15x-14}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$.
   3. Clear denominators: $15x-14=A(x-2)+B(x-1)$.
   4. Put $x=1$: $15-14=A(-1)\Rightarrow 1=-A\Rightarrow A=-1$. …

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7. 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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8. 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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9. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^3 - 1}{x^3 + x^2}\, dx = f(x) + \log(g(x)) + c$, $f(1) = 2$ and $g(-3) = \dfrac{3}{4}$, then $f(-2) + g(-2) =$ (A) $-\dfrac{9}{2}$ (B) $-\dfrac{1}{2}$ (C) $\dfrac{9}{4}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   A rational-function integral splits by partial fractions into a polynomial/rational part $f(x)$ plus a logarithmic part $\log(g(x))$; the two given data points pin down $f$ and $g$ exactly, letting us evaluate $f(-2)+g(-2)$. Answer: $-\tfrac12$.

   Concept and Intuition

   When an improper rational integrand is written $\frac{x^3-1}{x^3+x^2}$, polynomial division peels off the constant part, and partial fractions turn the remaining proper fraction into simple terms of the form $\frac{A}{x}, \frac{B}{x^2}, \frac{C}{x+1}$ whose antiderivatives are $A\log|x|$, $-B/x$, $C\log|x+1|$. Collecting all log terms into a single $\log(g(x))$ and all algebraic terms into $f(x)$ matches the form the question gives; the two numeric conditions are just there to confirm the constants (and resolve the sign inside the absolute value at negative $x$).

   Step-by-Step Solution

   1. Divide: $\dfrac{x^3-1}{x^3+x^2}=1-\dfrac{x^2+1}{x^3+x^2}=1-\dfrac{x^2+1}{x^2(x+1)}$.
   2. Partial fractions: $\dfrac{x^2+1}{x^2(x+1)}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x+1}$. Clearing denominators: $x^2+1=Ax(x+1)+B(x+1)+Cx^2$. Setting $x=0$: $B=1$. Setting $x=-1$: $2=C$. Matching $x^2$ coefficients: $1=A+C\Rightarrow A=-1$.
   3. So $\dfrac{x^2+1}{x^2(x+1)}=-\dfrac1x+\dfrac1{x^2}+\dfrac2{x+1}$, and the integrand is $1+\dfrac1x-\dfrac1{x^2}-\dfrac2{x+1}$.
   4. Integrate term by term: $\displaystyle\int\left(1+\frac1x-\frac1{x^2}-\frac{2}{x+1}\right)dx=x+\log|x|+\frac1x-2\log|x+1|+c=\left(x+\frac1x\right)+\log\left|\frac{x}{(x+1)^2}\right|+c$. …

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10. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.If $\dfrac{x^4}{(x-1)(x-2)} = f(x) + \dfrac{A}{x-1} + \dfrac{B}{x-2}$, then (A) $f(x) = x^2 - 3x + 7$ (B) $f(x) = x^2 + 3x + 7$ (C) $A + B = 17$ (D) $A - B = -18$

    ›Reveal solutionSolution

    Dividing $x^4$ by $(x-1)(x-2)$ gives quotient $f(x)=x^2+3x+7$ with remainder $15x-14$, and partial-fractioning the remainder gives $A=-1,\,B=16$ — so only the $f(x)$ option is correct.

    Concept and Intuition

    When a rational function's numerator degree exceeds the denominator's, you must first do polynomial long division to extract the polynomial part $f(x)$ before decomposing the proper-fraction remainder into partial fractions. Skipping the division step and partial-fractioning directly would be invalid since $\frac{x^4}{(x-1)(x-2)}$ isn't a proper fraction.

    Step-by-Step Solution

    1. Divide $x^4$ by $x^2-3x+2$ (the denominator's product):
       • $x^4 \div x^2 = x^2$; $x^2(x^2-3x+2)=x^4-3x^3+2x^2$; subtract: $3x^3-2x^2$.
       • $3x^3 \div x^2 = 3x$; $3x(x^2-3x+2)=3x^3-9x^2+6x$; subtract: $7x^2-6x$.
       • $7x^2 \div x^2 = 7$; $7(x^2-3x+2)=7x^2-21x+14$; subtract: $15x-14$.
    2. So $x^4 = (x^2-3x+2)(x^2+3x+7) + (15x-14)$, giving $f(x)=x^2+3x+7$ and remainder $15x-14$.
    3. Decompose $\dfrac{15x-14}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}$: so $15x-14=A(x-2)+B(x-1)$.
    4. At $x=1$: $15-14=1=A(-1) \Rightarrow A=-1$. …

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11. 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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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.If $\dfrac{2x^4 - 3x^2 + 4}{(x^2+1)(x^2+2)} = a + \dfrac{px+q}{x^2+1} + \dfrac{mx+n}{x^2+2}$, then $\dfrac{n}{q} =$ (A) $p + m - a$ (B) $\dfrac{p+m}{a}$ (C) $\dfrac{a}{p+m}$ (D) $p + m + a$

    ›Reveal solutionSolution

    Clearing denominators in the partial-fraction identity and matching coefficients of each power of $x$ pins down $a,p,q,m,n$ exactly; the ratio $n/q$ works out to $-2$, which matches option (A)'s expression $p+m-a$.

    Concept and Intuition

    When a proper (or, as here, degree-equal) rational function is split into partial fractions with irreducible quadratic denominators, the standard method is to multiply through by the common denominator and equate coefficients of like powers of $x$ on both sides — this converts the identity into a system of linear equations in the unknown constants.

    Step-by-Step Solution

    1. Multiply both sides by $(x^2+1)(x^2+2)$: $2x^4-3x^2+4 = a(x^2+1)(x^2+2)+(px+q)(x^2+2)+(mx+n)(x^2+1)$.
    2. Expand $a(x^2+1)(x^2+2)=a(x^4+3x^2+2)$, $(px+q)(x^2+2)=px^3+qx^2+2px+2q$, $(mx+n)(x^2+1)=mx^3+nx^2+mx+n$.
    3. Collect by power of $x$:

    • $x^4$: $a=2$
    • $x^3$: $p+m=0$
    • $x^2$: $3a+q+n=-3\Rightarrow q+n=-3-6=-9$
    • $x^1$: $2p+m=0$
    • $x^0$: $2a+2q+n=4\Rightarrow 2q+n=4-4=0$

    4. From $p+m=0$ and $2p+m=0$: subtracting gives $p=0$, so $m=0$. …

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