Q.Integrate the following function: $\frac{1}{x^2 - 9}$

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 $\displaystyle\int\frac{x-1}{(x-2)(x-3)}dx = -\log|x-2| + 2\log|x-3| + c$.
   5. Combine: $2\log|x-3|-\log|x-2| = \log\dfrac{|x-3|^2}{|x-2|} = \log\left|\dfrac{(x-3)^2}{x-2}\right|$.

   Common Mistakes

   • Getting the sign of $A$ wrong (it is $-1$, not $+1$), which flips the final combined-log form and matches the wrong option.
   • Combining logs incorrectly, e.g. writing $\log|x-3|^2+\log|x-2|$ instead of the correct difference.

   ✓Final answer

   The correct option is (D) — $\log\left|\dfrac{(x-3)^2}{x-2}\right|+c$.

   ANSWER: D

2. 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 $\displaystyle\int\left(-\dfrac{1/3}{x-1}+\dfrac{4/3}{x-4}\right)dx = -\dfrac13\log|x-1| + \dfrac43\log|x-4| + c$.
   7. Factor out $\dfrac13$: $= \dfrac13\left[4\log|x-4| - \log|x-1|\right]+c = \dfrac13\log\dfrac{|x-4|^4}{|x-1|}+c = \dfrac13\log\dfrac{(x-4)^4}{|x-1|}+c$ (since $(x-4)^4$ is always non-negative, the absolute value on it is unnecessary).

   Common Mistakes

   • Sign errors when solving for $A$ and $B$ using the cover-up/substitution method.
   • Combining the two log terms incorrectly (e.g. adding instead of subtracting, or mismatching which term gets the power of 4).

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{3}\log\dfrac{(x-4)^4}{|x-1|} + c$.

   ANSWER: A

3. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If $\dfrac{x^2+3}{x^4+2x^2+9} = \dfrac{Ax+B}{x^2+ax+b} + \dfrac{Cx+D}{x^2+cx+b}$ then $aA + bB + cC + D =$ (A) $1$ (B) $0$ (C) $-1$ (D) $2$

   ›Reveal solutionSolution

   Factor the quartic denominator into two quadratics, split the fraction, solve for the constants by matching coefficients, and evaluate the requested combination. Answer: $2$.

   Concept and Intuition

   The quartic $x^4+2x^2+9$ has the classic Sophie-Germain-like factorization $(x^2+px+q)(x^2-px+q)=x^4+(2q-p^2)x^2+q^2$; matching $q^2=9,\ 2q-p^2=2$ gives $q=3,\ p=2$. Once the denominators are known, comparing coefficients of a polynomial identity pins down $A,B,C,D$ uniquely.

   Step-by-Step Solution

   1. Factor: $x^4+2x^2+9=(x^2+2x+3)(x^2-2x+3)$, so $a=2,\ b=3,\ c=-2$ (matching $(x^2+ax+b)(x^2+cx+b)$).
   2. Write $\dfrac{x^2+3}{(x^2+2x+3)(x^2-2x+3)}=\dfrac{Ax+B}{x^2+2x+3}+\dfrac{Cx+D}{x^2-2x+3}$.
   3. Combine and match numerators: $(Ax+B)(x^2-2x+3)+(Cx+D)(x^2+2x+3)=x^2+3$. Expanding and comparing coefficients of $x^3,x^2,x^1,x^0$ gives $A+C=0$, $D=B$, and solving the system yields $A=0,\ C=0,\ B=D=\dfrac12$.
   4. Verify: $\dfrac{1/2}{x^2+2x+3}+\dfrac{1/2}{x^2-2x+3}=\dfrac{(x^2-2x+3)+(x^2+2x+3)}{2(x^4+2x^2+9)}=\dfrac{2x^2+6}{2(x^4+2x^2+9)}=\dfrac{x^2+3}{x^4+2x^2+9}$ ✓.
   5. Compute $aA+bB+cC+D = 2(0)+3\left(\tfrac12\right)+(-2)(0)+\tfrac12 = \tfrac32+\tfrac12=2$.

   Common Mistakes

   • Missing the correct factorization of the quartic (guessing $(x^2+3)^2-\dots$ style factorizations that don't match).
   • Sign slip when identifying which linear denominator coefficient is $a$ and which is $c$.

   ✓Final answer

   The correct option is (D) — $2$.

   ANSWER: D

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).
   7. Total: $\frac23\log|x-1|-\frac13\log(x^2+x+1)+c=\frac13\left[2\log|x-1|-\log(x^2+x+1)\right]+c=\frac13\log\dfrac{(x-1)^2}{x^2+x+1}+c$.

   Common Mistakes

   • Forgetting to square $(x-1)$ when combining the $\frac23\log|x-1|$ term into a single logarithm.
   • Missing that the quadratic-factor numerator has no residual arctan piece here (since $C$ makes it a pure multiple of the derivative).

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{3}\log\left(\dfrac{(x-1)^2}{x^2+x+1}\right)+c$.

   ANSWER: B

5. 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

   • Getting the log argument upside down: $\log\left|\dfrac{x+1}{x-1}\right|$ vs $\log\left|\dfrac{x-1}{x+1}\right|$ differ by an overall sign, which flips which option matches — always verify via $\int\frac{dx}{x^2-1}=\frac12\log\left|\frac{x-1}{x+1}\right|$ specifically (not the $a^2-x^2$ version).
   • Sign error on the $\tan^{-1}x$ term.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{4}\log\left|\dfrac{x-1}{x+1}\right|+\dfrac{1}{2}\mathrm{Tan}^{-1}x+c$.

   ANSWER: B

6. 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 $x=-2$: $5=0+0+C(-4)(5)\Rightarrow -20C=5\Rightarrow C=-\tfrac14$.
   5. Compare $x^2$ coefficients: $A+2B-2C=2\Rightarrow A+0.5+0.5=2\Rightarrow A=1$.
   6. $6A+7B-5C=6(1)+7(0.25)-5(-0.25)=6+1.75+1.25=9$.

   Common Mistakes

   • Sign slip when substituting $x=-2$ (the $(x-2)$ factor becomes $-4$, easy to mishandle).
   • Forgetting to verify with the constant-term or $x$-coefficient equation as a consistency check.

   ✓Final answer

   The correct option is (A) — $9$.

   ANSWER: A

7. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If $\dfrac{1}{(3-5x)(2+3x)}=\dfrac{A}{3-5x}+\dfrac{B}{2+3x}$ then $A+B=$ (A) $\dfrac{7}{19}$ (B) $\dfrac{8}{19}$ (C) $\dfrac{9}{19}$ (D) $\dfrac{10}{19}$

   ›Reveal solutionSolution

   Standard partial fractions: clearing denominators and plugging in the roots of each factor gives $A=5/19$, $B=3/19$, so $A+B=8/19$.

   Concept and Intuition

   For a proper rational function with distinct linear factors in the denominator, each constant is found by substituting the value of $x$ that zeroes out the OTHER factor.

   Step-by-Step Solution

   1. Write $\dfrac{1}{(3-5x)(2+3x)}=\dfrac{A}{3-5x}+\dfrac{B}{2+3x}$.
   2. Multiply through: $1=A(2+3x)+B(3-5x)$.
   3. Set $3-5x=0 \Rightarrow x=3/5$: $1=A(2+3\cdot3/5)=A(2+9/5)=A\cdot\dfrac{19}{5}\Rightarrow A=\dfrac{5}{19}$.
   4. Set $2+3x=0\Rightarrow x=-2/3$: $1=B(3-5\cdot(-2/3))=B(3+10/3)=B\cdot\dfrac{19}{3}\Rightarrow B=\dfrac{3}{19}$.
   5. $A+B=\dfrac{5}{19}+\dfrac{3}{19}=\dfrac{8}{19}$.

   Common Mistakes

   • Sign errors when solving for the zeroing value of $x$ for each factor.
   • Forgetting to keep $A,B$ as separate fractions with the same denominator (19) before adding.

   ✓Final answer

   The correct option is (B) — $\dfrac{8}{19}$.

   ANSWER: B

8. 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: $\displaystyle\int\left(1+\frac1x-\frac1{x^2}-\frac{2}{x+1}\right)dx=x+\log|x|+\frac1x-2\log|x+1|+c=\left(x+\frac1x\right)+\log\left|\frac{x}{(x+1)^2}\right|+c$.
   5. So $f(x)=x+\dfrac1x$; check $f(1)=1+1=2$ ✓ (matches the given condition exactly, with no extra constant needed).
   6. $g(x)$ must equal $\left|\dfrac{x}{(x+1)^2}\right|$ so that $\log(g(x))=\log\left|\dfrac{x}{(x+1)^2}\right|$; check $g(-3)=\dfrac{3}{4}$ ✓ (matches exactly).
   7. So $g(x)=\dfrac{|x|}{(x+1)^2}$ for $x<0$ (i.e. $g(x)=\dfrac{-x}{(x+1)^2}$ there).
   8. $f(-2)=-2+\dfrac{1}{-2}=-2-\dfrac12=-\dfrac52$. $g(-2)=\dfrac{|-2|}{(-1)^2}=\dfrac{2}{1}=2$.
   9. $f(-2)+g(-2)=-\dfrac52+2=-\dfrac12$.

   Common Mistakes

   • Forgetting the $|x|$/sign subtlety in $g(x)$ for negative $x$, which would give the wrong sign in $g(-2)$.
   • Arithmetic slip in the partial-fraction coefficients.

   ✓Final answer

   The correct option is (B) — $-\dfrac{1}{2}$.

   ANSWER: B

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 $x=-2$: $9 = C(-2-1) = -3C \Rightarrow C=-3$.
   4. Compare coefficients of $x^2$ on both sides: RHS gives $A+B$ (from $A(x+2)^2$ contributing $Ax^2$ and $B(x-1)(x+2)$ contributing $Bx^2$), LHS has 0. So $A+B=0 \Rightarrow B=-1$.
   5. Check constant term: $4A-2B-C = 4(1)-2(-1)-(-3)=4+2+3=9$ ✓, confirming the values.
   6. $A-B-C = 1-(-1)-(-3) = 1+1+3 = 5$.

   Common Mistakes

   • Forgetting the extra $\dfrac{C}{(x+2)^2}$ term for the repeated factor.
   • Sign errors when substituting $x=-2$ or $x=1$.

   ✓Final answer

   The correct option is (B) — 5.

   ANSWER: B

10. 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 $t=-\tfrac12$: $C=-\tfrac23$.
    6. So $-\displaystyle\int\left[\frac{1/6}{1-t}+\frac{1/2}{1+t}-\frac{2/3}{1+2t}\right]dt = \frac16\log|1-t|+\frac12\log|1+t|-\frac23\log|1+2t|+c$ (the sign flips on the $1/(1-t)$ term because $\int\frac{dt}{1-t}=-\log|1-t|$, and the overall integral carries a leading minus sign).
    7. Substitute back $t=\cos x$: $\dfrac16\log|1-\cos x|+\dfrac12\log|1+\cos x|-\dfrac23\log|1+2\cos x|+c$.

    Common Mistakes

    • Missing the overall minus sign introduced by $dt=-\sin x\,dx$, which flips the sign of every term if mishandled.
    • Misassigning the partial-fraction coefficients (double-check by plugging $t=0$: $A+B+C$ should equal $1$).

    ✓Final answer

    The correct option is (C) — $\frac{1}{6}\log|1-\cos x| + \frac{1}{2}\log|1+\cos x| - \frac{2}{3}\log|1+2\cos x| + c$.

    ANSWER: C

11. 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 $\dfrac{3x-4}{(x+4)(x-1)} = \dfrac{16}{5(x+4)} - \dfrac{1}{5(x-1)}$.
    7. Therefore $\dfrac{x^2}{x^2+3x-4} = 1 - \dfrac{16}{5(x+4)} + \dfrac{1}{5(x-1)} = 1+\dfrac{-16}{5(x+4)}+\dfrac{1}{5(x-1)}$.

    Common Mistakes

    • Forgetting the initial polynomial division step (since numerator and denominator have equal degree), and trying to decompose the improper fraction directly.

    ✓Final answer

    The correct option is (A) — $1+\dfrac{-16}{5(x+4)}+\dfrac1{5(x-1)}$.

    ANSWER: A

12. 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 $=(A+B)(2)+3C = 2A+2B+3C$.

    $0 = 2A + 2\left(\frac{13}{19}\right)+3\left(-\frac{22}{19}\right) = 2A+\frac{26-66}{19}=2A-\frac{40}{19}\Rightarrow A=\frac{20}{19}.$

    5. So $\dfrac{Ax+B}{x^2+2}=\dfrac{\frac{20}{19}x+\frac{13}{19}}{x^2+2}=\dfrac{20x+13}{19(x^2+2)}$, and separately $\dfrac{C}{3x-1}=\dfrac{-22}{19(3x-1)}$.
    6. Matching against the options, the fraction $\dfrac{20x+13}{19(x^2+2)}$ is exactly option (C).

    Common Mistakes

    • Sign slips when solving the linear system for $A,B,C$ — always re-verify by plugging a value back into the original equation.
    • Confusing the sign or the denominator constant on the $C/(3x-1)$ term with the $(Ax+B)/(x^2+2)$ term when matching to the options — here the two "distractor" options (A) and (D) are actually built from the other partial fraction ($C$-term), not this one.

    ✓Final answer

    The correct option is (C) — $\dfrac{20x+13}{19(x^2+2)}$.

    ANSWER: C