Q.If $\int \dfrac{dx}{(x+2)(x^2+1)} = a\log|1+x^2| + b\tan^{-1}x + \dfrac{1}{5}\log|x+2| + C$, then
(A) $a=-\dfrac{1}{10},\ b=-\dfrac{2}{5}$
(B) $a=\dfrac{1}{10},\ b=-\dfrac{2}{5}$
(C) $a=-\dfrac{1}{10},\ b=\dfrac{2}{5}$
(D) $a=\dfrac{1}{10},\ b=\dfrac{2}{5}$

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 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\dfrac{A}{(x+2)^2}$ is one of the partial fractions of $\dfrac{x^2+3x+5}{(2x+1)(x+2)^2}$, then $A =$ (A) 2 (B) 1 (C) -2 (D) -1

   ›Reveal solutionSolution

   Using the cover-up (Heaviside) method at the repeated root $x=-2$ gives $A=-1$ directly, no full partial-fraction expansion needed.

   Concept and Intuition

   For a repeated linear factor $(x+2)^2$ in the denominator, the partial fraction decomposition has a term $\dfrac{A}{(x+2)^2}$ whose coefficient can be found by the cover-up method: multiply both sides by $(x+2)^2$, which cancels that factor from the denominator entirely, then substitute $x=-2$ into what remains (the other terms of the decomposition vanish or become finite/zero at this special substitution for the squared-factor coefficient specifically).

   Step-by-Step Solution

   1. Write $\dfrac{x^2+3x+5}{(2x+1)(x+2)^2}=\dfrac{P}{2x+1}+\dfrac{Q}{x+2}+\dfrac{A}{(x+2)^2}$.
   2. Multiply both sides by $(x+2)^2$: $\dfrac{x^2+3x+5}{2x+1} = P\cdot\dfrac{(x+2)^2}{2x+1}+Q(x+2)+A$.
   3. Substitute $x=-2$ (this kills the $P$ and $Q$ terms since they still carry a factor of $(x+2)$): $A=\dfrac{x^2+3x+5}{2x+1}\bigg|_{x=-2}$. …

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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 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{2x+5}{(x-1)(x+1)(x+4)(x+6)}\, dx = \dfrac{1}{10}\log\left(\dfrac{f(x)}{g(x)}\right) + c$ and $\dfrac{f(-2)}{g(-2)} = 6$, then $\dfrac{f(10)}{g(10)} =$ (A) $\dfrac{72}{77}$ (B) $\dfrac{144}{75}$ (C) $\dfrac{55}{63}$ (D) $\dfrac{70}{59}$

   ›Reveal solutionSolution

   Grouping the quartic denominator via $u=x^2+5x$ turns it into $(u-6)(u+4)$, whose partial fractions give $f(x)=(x-1)(x+6)$ and $g(x)=(x+1)(x+4)$; evaluating at $x=10$ gives $144/154=72/77$.

   Concept and Intuition

   $(x-1)(x+6)=x^2+5x-6$ and $(x+1)(x+4)=x^2+5x+4$ share the same quadratic core $x^2+5x$. Substituting $u=x^2+5x$ (whose derivative $2x+5$ is exactly the numerator) collapses the quartic denominator into a simple product $(u-6)(u+4)$, reducing the problem to a standard $\int \frac{du}{(u-a)(u-b)}$ form.

   Step-by-Step Solution

   1. Let $u=x^2+5x \Rightarrow du=(2x+5)dx$. Then

   $(x-1)(x+1)(x+4)(x+6)=(u-6)(u+4)$

   since $(x-1)(x+6)=x^2+5x-6=u-6$ and $(x+1)(x+4)=x^2+5x+4=u+4$.

   2. The integral becomes

   $\int\frac{du}{(u-6)(u+4)}=\frac{1}{10}\int\left(\frac{1}{u-6}-\frac{1}{u+4}\right)du=\frac{1}{10}\log\left|\frac{u-6}{u+4}\right|+c$

   3. So $\dfrac{f(x)}{g(x)}=\dfrac{u-6}{u+4}=\dfrac{(x-1)(x+6)}{(x+1)(x+4)}$. …

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5. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.Given $\dfrac{3x-2}{(x+1)^2(x+3)} = \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2} + \dfrac{C}{x+3}$ then $4A + 2B + 4C$ (A) $5$ (B) $-5$ (C) $-3$ (D) $3$

   ›Reveal solutionSolution

   Standard partial-fraction cover-up plus a coefficient match pins down $A,B,C$, giving $4A+2B+4C=-5$.

   Concept and Intuition

   For a repeated linear factor $(x+1)^2$ together with a simple factor $(x+3)$, the cover-up (Heaviside) method quickly gives $B$ and $C$ by substituting the roots that make each factor vanish. The remaining constant $A$ is then found by matching a coefficient (here, the $x^2$ coefficient, since the numerator on the left has no $x^2$ term).

   Step-by-Step Solution

   1. Clear denominators: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$.
   2. Set $x=-1$ (kills the $A$ and $C$ terms): $3(-1)-2 = B(-1+3) \Rightarrow -5 = 2B \Rightarrow B=-\dfrac{5}{2}$.
   3. Set $x=-3$ (kills the $A$ and $B$ terms): $3(-3)-2 = C(-3+1)^2 \Rightarrow -11 = 4C \Rightarrow C=-\dfrac{11}{4}$.
   4. Match the coefficient of $x^2$ on both sides (LHS has none): $0 = A + C \Rightarrow A = \dfrac{11}{4}$. …

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6. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $\dfrac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{Cx-3}{x^2+2}$, then $A+B+C=$ (A) $7$ (B) $5$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   Clearing denominators and matching coefficients (using $x=1$ to isolate $B$ first) gives $A=0,\ B=2,\ C=0$, so $A+B+C=2$.

   Concept and Intuition

   This is a standard partial-fractions decomposition. The repeated linear factor $(x-1)^2$ lets us find $B$ instantly by substituting $x=1$ directly into the cleared equation (this kills every term except the one multiplying $B$). The remaining coefficients $A,C$ are then found by matching powers of $x$.

   Step-by-Step Solution

   1. Clear denominators: $-x^2+6x+1=A(x-1)(x^2+2)+B(x^2+2)+(Cx-3)(x-1)^2$.
   2. Set $x=1$: LHS $=-1+6+1=6$; RHS $=0+B(3)+0=3B$. So $B=2$.
   3. Expand the RHS fully with $B=2$: $A(x^3-x^2+2x-2)+2x^2+4+(Cx-3)(x^2-2x+1)$. $(Cx-3)(x^2-2x+1)=Cx^3-2Cx^2+Cx-3x^2+6x-3$.
   4. Collect by power of $x$: $x^3$: $A+C$ $x^2$: $-A+2-2C-3=-A-2C-1$ $x^1$: $2A+C+6$ $x^0$: $-2A+4-3=-2A+1$
   5. Match to LHS coefficients ($0,-1,6,1$ for $x^3,x^2,x,1$): $x^3$: $A+C=0$ …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\dfrac{3x+1}{(x-1)(x^2+2)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+2}$, then $5(A-B)=$ (A) $A+C$ (B) $8C$ (C) $C+8$ (D) $\dfrac{C}{8}$

   ›Reveal solutionSolution

   This tests partial-fraction decomposition and coefficient comparison. Answer: $5(A-B)=8C$.

   Concept and Intuition

   To find $A,B,C$ in a partial fraction decomposition, clear denominators to get a polynomial identity, then either substitute convenient values of $x$ (like the root of the linear factor) or compare coefficients of like powers of $x$.

   Step-by-Step Solution

   1. Clear denominators: $3x+1 = A(x^2+2)+(Bx+C)(x-1)$.
   2. Substitute $x=1$ (kills the $(Bx+C)(x-1)$ term): $3(1)+1 = A(1+2) \Rightarrow 4=3A \Rightarrow A=\dfrac43$.
   3. Expand the right side: $Ax^2+2A+Bx^2-Bx+Cx-C = (A+B)x^2+(C-B)x+(2A-C)$.
   4. Compare coefficient of $x^2$ (LHS has none): $A+B=0\Rightarrow B=-A=-\dfrac43$.
   5. Compare coefficient of $x^1$: $C-B=3\Rightarrow C=3+B=3-\dfrac43=\dfrac53$.
   6. Check constant term: $2A-C = \dfrac83-\dfrac53=1$ ✓, consistent. …

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8. 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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9. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $\dfrac{9}{(x-1)(x+2)^2} = \dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$ then $A-B-C$ is equal to (A) 3 (B) 5 (C) $-1$ (D) 0

   ›Reveal solutionSolution

   This is a standard partial-fractions problem; plugging in convenient roots quickly isolates each constant, giving $A-B-C=5$.

   Concept and Intuition

   For a repeated linear factor $(x+2)^2$, the partial fraction decomposition needs both a $\dfrac{B}{x+2}$ and a $\dfrac{C}{(x+2)^2}$ term. The fastest way to find the constants is to clear denominators and substitute the roots of the linear factors directly (this instantly kills all but one term).

   Step-by-Step Solution

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

   $9 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)$

   2. Put $x=1$: $9 = A(3)^2 = 9A \Rightarrow A=1$.
   3. Put $x=-2$: $9 = C(-2-1) = -3C \Rightarrow C=-3$. …

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10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.If $\dfrac{x+1}{(x-1)^2(x^2+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2+1}$, then $\sqrt{3A^2+4D^2+5C^2+B^2} =$ (A) $\dfrac{3}{2}$ (B) $\dfrac{1}{2}$ (C) 1 (D) 2

    ›Reveal solutionSolution

    Solving the partial-fraction decomposition gives $A=-\tfrac12, B=1, C=\tfrac12, D=-\tfrac12$; substituting into $\sqrt{3A^2+4D^2+5C^2+B^2}$ gives $2$.

    Concept and Intuition

    A rational function with a repeated linear factor $(x-1)^2$ and an irreducible quadratic factor $(x^2+1)$ decomposes as $\dfrac A{x-1}+\dfrac B{(x-1)^2}+\dfrac{Cx+D}{x^2+1}$. Clearing denominators and matching coefficients (or plugging convenient values of $x$) pins down all four constants.

    Step-by-Step Solution

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

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

    2. Plug $x=1$: LHS $=2$. RHS $= 0 + B(1+1) + 0 = 2B$. So $B=1$.
    3. Expand each term:
       • $A(x-1)(x^2+1) = A(x^3-x^2+x-1)$.
       • $B(x^2+1) = x^2+1$ (using $B=1$).
       • $(Cx+D)(x-1)^2 = (Cx+D)(x^2-2x+1) = Cx^3+(-2C+D)x^2+(C-2D)x+D$.
    4. Collect coefficients and match with $x+1 = 0\cdot x^3+0\cdot x^2+1\cdot x+1$:
       • $x^3$: $A+C=0$.
       • $x^2$: $-A+1-2C+D=0$.
       • $x^1$: $A+C-2D=1$.
       • $x^0$: $-A+1+D=1$.
    5. From $x^3$: $C=-A$. Substitute into $x^1$ equation: $A-A-2D=1 \Rightarrow D=-\tfrac12$.
    6. From $x^0$: $-A+D=0 \Rightarrow A=D=-\tfrac12$, hence $C=-A=\tfrac12$.
    7. (Verify $x^2$ equation: $-(-\tfrac12)+1-2(\tfrac12)+(-\tfrac12) = \tfrac12+1-1-\tfrac12=0$ ✓.) …

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11. 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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12. 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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