Q.Find $\int \dfrac{x^4\, dx}{(x-1)(x^2+1)}$

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

   • 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

2. 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

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 $\displaystyle\int\left(-\dfrac{1/3}{x-1}+\dfrac{4/3}{x-4}\right)dx = -\dfrac13\log|x-1| + \dfrac43\log|x-4| + c$.
   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

4. 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 $1+\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+3}$, with $A,B,C,D$ real (here in fact $A=C=0$ since the numerator is even, but the form required allows any reals, which is exactly what option (D) states).

   Common Mistakes

   • Jumping straight to $\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+3}$ (option A) without noticing the numerator/denominator degrees are equal, which misses the required "+1".
   • Assuming $A,B,C,D$ must all be nonzero — the form just needs them to be real; some can turn out to be zero once solved.

   ✓Final answer

   The correct option is (D) — $1 + \dfrac{Ax+B}{x^2+1} + \dfrac{Cx+D}{x^2+3}$ for some $A,B,C,D\in\mathbb{R}$.

   ANSWER: D

5. 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

6. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.If $\dfrac{x}{(x-1)(x^2+1)^2} = \dfrac{1}{4}\left[\dfrac{1}{x-1} - \dfrac{x+1}{x^2+1}\right]+y$, then $y =$ (A) $\dfrac{1}{2}\left[\dfrac{1-x}{(x^2+1)^2}\right]$ (B) $\dfrac{1+x}{3(x^2+1)^2}$ (C) $\dfrac{1-x}{(x^2-1)^2}$ (D) $\dfrac{1+x}{(x^2+1)^2}$

   ›Reveal solutionSolution

   This tests partial fraction decomposition with a repeated irreducible quadratic factor, then matching the remaining unaccounted term to $y$.

   Concept and Intuition

   For a denominator $(x-1)(x^2+1)^2$, the full decomposition has the form $\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}$. The problem already gives the first two pieces combined as $\tfrac14\left[\tfrac{1}{x-1}-\tfrac{x+1}{x^2+1}\right]$, so $y$ must be exactly the third piece.

   Step-by-Step Solution

   1. Write $\dfrac{x}{(x-1)(x^2+1)^2} = \dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}$.
   2. Multiply through: $x = A(x^2+1)^2 + (Bx+C)(x-1)(x^2+1) + (Dx+E)(x-1)$.
   3. Set $x=1$: $1 = 4A \Rightarrow A=\tfrac14$.
   4. Expand and match coefficients of $x^4,x^3,x^2,x^1,x^0$: this yields $B=-\tfrac14,\ C=-\tfrac14,\ D=-\tfrac12,\ E=\tfrac12$.
   5. So the full decomposition is $\dfrac{1/4}{x-1} - \dfrac14\cdot\dfrac{x+1}{x^2+1} + \dfrac{(1-x)/2}{(x^2+1)^2}$.
   6. Comparing to the given equation, $y = \dfrac{1}{2}\cdot\dfrac{1-x}{(x^2+1)^2}$.
   7. Numerical check at $x=2$: LHS $=2/[(1)(25)]=0.08$; RHS $=\tfrac14(1)-\tfrac14\cdot\tfrac35+\tfrac{(1-2)}{2\cdot25}=0.25-0.15-0.02=0.08$. ✓

   Common Mistakes

   • Sign errors when equating the $x^0$ (constant) and $x^1$ coefficients — always cross-check with a numeric substitution as done above.

   ✓Final answer

   The correct option is (A) — $\dfrac12\left[\dfrac{1-x}{(x^2+1)^2}\right]$.

   ANSWER: A

7. 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 $C=-2E=-\dfrac12$.
   • $x^1$: $(-A+B)+(D-C)=1$. Since $-A+B=0$, this gives $D=1+C=1-\dfrac12=\dfrac12$.
   • $x^0$ (consistency check): $-B-D+E = \dfrac14-\dfrac12+\dfrac14=0$ ✓.

   5. So $A=-\dfrac14$, $B=-\dfrac14$, $C=-\dfrac12$, $D=\dfrac12$, $E=\dfrac14$.
   6. Compute the required combination:

   $A+B-C+2D = \left(-\frac14\right)+\left(-\frac14\right)-\left(-\frac12\right)+2\left(\frac12\right) = -\frac12+\frac12+1 = 1.$

   Common Mistakes

   • Sign errors when expanding $(x^2+1)(x-1)$ or when matching the $x^2$/$x^1$ coefficients.
   • Forgetting to use the fast root-substitution trick for $E$ and instead trying to solve a 5×5 system from scratch.
   • Arithmetic slip in the final combination — easy to drop a sign on $-C$ since $C$ itself is negative.

   ✓Final answer

   The correct option is (B) — 1.

   ANSWER: B

8. 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 $A_2=1,A_4=-1$: $0=-A_1+A_3 \Rightarrow A_1=A_3$.
   5. Set $x=2$: $8=A_1(1)(9)+A_2(9)+A_3(3)(1)+A_4(1)=9A_1+9A_2+3A_3+A_4$. Substitute $A_2=1,A_4=-1,A_3=A_1$: $8=9A_1+9+3A_1-1=12A_1+8 \Rightarrow A_1=0$, so $A_3=0$.
   6. Sum: $A_1+A_2+A_3+A_4=0+1+0-1=0$.

   Common Mistakes

   • Only using the two root substitutions ($x=1,-1$) and forgetting extra points are needed to separate $A_1$ from $A_3$.
   • Sign errors when expanding $(x-1)(x+1)^2$ and $(x+1)(x-1)^2$ at the chosen test points.

   ✓Final answer

   The correct option is (C) — $0$.

   ANSWER: C

9. 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

10. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.The partial fraction decomposition of $\dfrac{x^4 + 24x^2 + 28}{(x^2+1)^3}$ is (A) $\dfrac{1}{x^2+1} - \dfrac{22}{(x^2+1)^2} + \dfrac{5}{(x^2+1)^3}$ (B) $\dfrac{1}{x^2+1} + \dfrac{22}{(x^2+1)^2} + \dfrac{5}{(x^2+1)^3}$ (C) $\dfrac{1}{x^2+1} - \dfrac{22}{(x^2+1)^2} - \dfrac{5}{(x^2+1)^3}$ (D) $\dfrac{1}{x^2+1} + \dfrac{22}{(x^2+1)^2} - \dfrac{5}{(x^2+1)^3}$

    ›Reveal solutionSolution

    This is a partial-fraction problem made easy by substituting $t=x^2+1$ so the whole numerator becomes a polynomial in $t$. Answer: all three signs are $+$, with coefficients $1,22,5$.

    Concept and Intuition

    Since the denominator is a power of $(x^2+1)$ only, and the numerator is a polynomial purely in $x^2$, it's far simpler to substitute $t=x^2+1$ (so $x^2=t-1$) and rewrite the numerator as a polynomial in $t$, rather than solving for unknown constants $A,B,C$ by matching coefficients directly in $x$.

    Step-by-Step Solution

    1. Let $t = x^2+1$, so $x^2 = t-1$, and $x^4 = (x^2)^2 = (t-1)^2 = t^2-2t+1$.
    2. Substitute into the numerator: $x^4+24x^2+28 = (t^2-2t+1) + 24(t-1) + 28$.
    3. Expand: $t^2-2t+1+24t-24+28 = t^2 + 22t + 5$.
    4. So the expression becomes $\dfrac{t^2+22t+5}{t^3} = \dfrac{t^2}{t^3}+\dfrac{22t}{t^3}+\dfrac{5}{t^3} = \dfrac{1}{t}+\dfrac{22}{t^2}+\dfrac{5}{t^3}$.
    5. Substituting back $t=x^2+1$: $\dfrac{1}{x^2+1}+\dfrac{22}{(x^2+1)^2}+\dfrac{5}{(x^2+1)^3}$.

    Common Mistakes

    • Sign errors while expanding $(t-1)^2$ or combining the $24(t-1)$ term.
    • Trying full partial fractions with undetermined constants $A,B,C$ over $(x^2+1),(x^2+1)^2,(x^2+1)^3$ separately — much slower and error-prone than the $t$-substitution.

    ✓Final answer

    The correct option is (B) — $\dfrac{1}{x^2+1} + \dfrac{22}{(x^2+1)^2} + \dfrac{5}{(x^2+1)^3}$.

    ANSWER: B

11. 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: $-(-\tfrac12)+1-2(\tfrac12)+(-\tfrac12) = \tfrac12+1-1-\tfrac12=0$ ✓.)
    8. So $A=-\tfrac12,\ B=1,\ C=\tfrac12,\ D=-\tfrac12$.
    9. Compute: $3A^2 = 3\cdot\tfrac14=\tfrac34$; $4D^2=4\cdot\tfrac14=1$; $5C^2=5\cdot\tfrac14=\tfrac54$; $B^2=1$.

    $3A^2+4D^2+5C^2+B^2 = \tfrac34+1+\tfrac54+1 = 4.$

    10. $\sqrt{4}=2$.

    Common Mistakes

    • Sign errors expanding $(Cx+D)(x-1)^2$ — expand $(x-1)^2$ first, then distribute carefully.
    • Forgetting to double back and verify with the unused coefficient equation (a good consistency check).
    • Arithmetic slip combining the fractions at the end — keep everything over a common denominator of $4$.

    ✓Final answer

    The correct option is (D) — $2$.

    ANSWER: D

12. 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