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

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-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 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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5. 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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6. 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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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 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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9. 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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10. 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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11. 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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12. 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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