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

Showing the 12 most recent of 63 on this concept.

1. 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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2. 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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3. 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. …

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4. 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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5. 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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6. 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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7. 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 $A=2$. …

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8. 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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9. 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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10. 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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11. 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 $(Ax+B)(x^2-3)=Ax^3+Bx^2-3Ax-3B$. …

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12. 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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