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

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 …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $\dfrac{3x-4}{(x+4)(x-1)} = \dfrac{16}{5(x+4)} - \dfrac{1}{5(x-1)}$. …

   🔒 Start your 14-day free trial to see the full solution →
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 $\displaystyle\int\left(-\dfrac{1/3}{x-1}+\dfrac{4/3}{x-4}\right)dx = -\dfrac13\log|x-1| + \dfrac43\log|x-4| + c$. …

   🔒 Start your 14-day free trial to see the full solution →
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 $x=-2$: $5=0+0+C(-4)(5)\Rightarrow -20C=5\Rightarrow C=-\tfrac14$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $A_2=1,A_4=-1$: $0=-A_1+A_3 \Rightarrow A_1=A_3$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 $(Ax+B)(x^2-3)=Ax^3+Bx^2-3Ax-3B$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $1+\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+3}$, …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $y=-2$: $-2=C(-3)(-1)=3C\Rightarrow C=-\dfrac23$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Which of the following is a partial fraction of $\dfrac{-x^{2}+6x+13}{(3x+5)(x^{2}+4x+4)} =$ (A) $\dfrac{3}{3x+5} + \dfrac{-1}{x+2} + \dfrac{2}{(x+2)^{2}}$ (B) $\dfrac{2}{3x+5} + \dfrac{-1}{x+2} + \dfrac{3}{(x+2)^{2}}$ (C) $\dfrac{-1}{3x+5} + \dfrac{2}{x+2} + \dfrac{3}{(x+2)^{2}}$ (D) $\dfrac{3}{3x+5} + \dfrac{2}{x+2} + \dfrac{-1}{(x+2)^{2}}$

   ›Reveal solutionSolution

   Clearing denominators and plugging in the repeated-root value, the linear-factor root, and matching one coefficient gives $A=2,\ B=-1,\ C=3$.

   Concept and Intuition

   For a repeated linear factor, plugging in its root isolates the coefficient of the highest power of that factor directly; the simple linear factor's root isolates its own coefficient; any remaining coefficient is found by matching a convenient power of $x$.

   Step-by-Step Solution

   1. Write $-x^2+6x+13 = A(x+2)^2 + B(3x+5)(x+2) + C(3x+5)$.
   2. At $x=-2$: LHS $=-4-12+13=-3$; RHS $=C(3(-2)+5)=-C$. So $C=3$.
   3. At $x=-5/3$: LHS $=-25/9-10+13=2/9$; RHS $=A(x+2)^2=A(1/3)^2=A/9$. So $A=2$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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

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

    🔒 Start your 14-day free trial to see the full solution →
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 $B=D$: $4B=1\Rightarrow B=D=\tfrac14$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.If $\dfrac{3x^2+x+2}{(3x^2+x+4)(3x^2+x+1)} = \dfrac{Ax+B}{3x^2+x+4} + \dfrac{Cx+D}{3x^2+x+1}$, then $(A+B) + (C+D) =$ (A) $\dfrac{1}{3}$ (B) $\dfrac{2}{3}$ (C) $1$ (D) $\dfrac{3}{2}$

    ›Reveal solutionSolution

    Substituting $u=3x^2+x$ collapses the problem to an ordinary constant partial fraction in $u$, forcing $A=C=0$ and giving $(A+B)+(C+D)=1$.

    Concept and Intuition

    When a rational expression's numerator and both denominator factors are built from the same quadratic block $3x^2+x$ shifted by constants, it's really a partial-fraction problem in the single variable $u=3x^2+x$, not in $x$ directly. Recognizing this shortcut avoids a messy 4-unknown system in $x$.

    Step-by-Step Solution

    1. Let $u=3x^2+x$. The equation becomes $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{Ax+B}{u+4}+\dfrac{Cx+D}{u+1}$.
    2. Do ordinary partial fractions in $u$: $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{P}{u+4}+\dfrac{Q}{u+1}$.
    3. Cover-up at $u=-4$: $P = \dfrac{-4+2}{-4+1} = \dfrac{-2}{-3}=\dfrac23$. At $u=-1$: $Q=\dfrac{-1+2}{-1+4}=\dfrac{1}{3}$.
    4. So $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{2/3}{u+4}+\dfrac{1/3}{u+1}$, which is entirely $x$-independent in its numerators. …

    🔒 Start your 14-day free trial to see the full solution →