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

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 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

4. 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

5. 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 $x^3,x^2,x,1$): $x^3$: $A+C=0$ $x^0$: $-2A+1=1\Rightarrow A=0\Rightarrow C=0$ (from the $x^3$ equation). Check $x^2$: $-0-0-1=-1$ ✓. Check $x^1$: $0+0+6=6$ ✓ — all consistent.
   6. $A+B+C=0+2+0=2$.

   Common Mistakes

   • Forgetting the shortcut of plugging $x=1$ to find $B$ instantly, and instead trying to solve a full $3\times3$ system.
   • Sign errors expanding $(Cx-3)(x-1)^2$.

   ✓Final answer

   The correct option is (D) — $2$.

   ANSWER: D

6. 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

7. 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): $0 = A + C \Rightarrow A = \dfrac{11}{4}$.
   5. Check with the constant term: $3A+3B+C = \dfrac{33}{4}-\dfrac{15}{2}-\dfrac{11}{4} = -2$ ✓, matching the LHS constant $-2$.
   6. Compute $4A+2B+4C = 4\left(\dfrac{11}{4}\right) + 2\left(-\dfrac52\right) + 4\left(-\dfrac{11}{4}\right) = 11 - 5 - 11 = -5$.

   Common Mistakes

   • Forgetting the repeated factor contributes two separate terms $\dfrac{B}{x+1}$ and $\dfrac{B'}{(x+1)^2}$ style constants — here correctly split as $A/(x+1)+B/(x+1)^2$.
   • Sign slips substituting negative roots.

   ✓Final answer

   The correct option is (B) — $-5$.

   ANSWER: B

8. 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

9. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $\dfrac{x^3}{(2x-1)(x+2)(x-3)} = A + \dfrac{B}{2x-1} + \dfrac{C}{x+2} + \dfrac{D}{x-3}$ then $A =$ (A) $\dfrac{1}{2}$ (B) $\dfrac{-1}{50}$ (C) $\dfrac{-8}{25}$ (D) $\dfrac{27}{25}$

   ›Reveal solutionSolution

   As $x\to\infty$, the left side tends to $1/2$ (ratio of leading coefficients) while the right side tends to $A$; hence $A=1/2$.

   Concept and Intuition

   Here the numerator's degree (3) equals the denominator's degree (3), so a genuine partial-fraction decomposition needs a polynomial (constant, here) term $A$ in addition to the proper-fraction terms. The value of $A$ can be found quickly by comparing leading behaviour as $x\to\infty$.

   Step-by-Step Solution

   1. Expand the denominator: $(2x-1)(x+2)(x-3) = 2x^3-3x^2-11x+6$.
   2. As $x\to\infty$: $\dfrac{x^3}{2x^3-3x^2-11x+6}\to\dfrac{1}{2}$.
   3. On the right side, as $x\to\infty$, all the proper-fraction terms $\dfrac{B}{2x-1},\dfrac{C}{x+2},\dfrac{D}{x-3}\to0$, leaving just $A$.
   4. Therefore $A=\dfrac12$.

   Common Mistakes

   • Trying to solve for $A,B,C,D$ simultaneously via a full system rather than noticing $A$ is isolable from the leading-order behaviour alone.
   • Missing that $A$ is needed at all because the numerator/denominator degrees are equal.

   ✓Final answer

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

   ANSWER: A

10. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.If the equivalent partial fraction of $\frac{x^3}{(2x-1)(x+2)(x-3)}$ is of the form $A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$ then the value of $A+B+C =$ (A) $-8/25$ (B) $4/25$ (C) $-1/50$ (D) $1/2$

    ›Reveal solutionSolution

    This is an improper partial fraction (numerator degree = denominator degree), so there's a constant term $A$ found from the leading behaviour, and $B,C$ are found by the standard cover-up (root-substitution) method — giving $A+B+C=4/25$.

    Concept and Intuition

    When the degree of the numerator equals the degree of the denominator, ordinary partial fractions leave a nonzero polynomial part (here, just a constant $A$, since both degrees are 3 and the denominator's leading coefficient is 2 — so $A$ equals the ratio of leading coefficients, $1/2$). The remaining proper-fraction coefficients ($B$, $C$, $D$) are then found efficiently using the "cover-up" trick: multiply through by the denominator and substitute each root of a linear factor to instantly isolate that factor's coefficient.

    Step-by-Step Solution

    1. As $x\to\infty$, $\dfrac{x^3}{(2x-1)(x+2)(x-3)} \to \dfrac{x^3}{2x^3} = \dfrac12$, so the constant part is $A=\dfrac12$.
    2. Multiply both sides by $(2x-1)(x+2)(x-3)$: $x^3 = A(2x-1)(x+2)(x-3) + B(x+2)(x-3) + C(2x-1)(x-3) + D(2x-1)(x+2)$.
    3. Set $x=\tfrac12$ (kills the $A$, $C$, $D$ terms): $\left(\tfrac12\right)^3 = B\left(\tfrac12+2\right)\left(\tfrac12-3\right) = B(2.5)(-2.5)=-6.25B$. $\tfrac18 = -\tfrac{25}{4}B \Rightarrow B = -\tfrac{1}{50}$.
    4. Set $x=-2$ (kills $A$, $B$, $D$): $(-2)^3 = C(2(-2)-1)(-2-3) = C(-5)(-5)=25C \Rightarrow -8=25C \Rightarrow C=-\tfrac{8}{25}$.
    5. Sum: $A+B+C = \tfrac12 - \tfrac1{50} - \tfrac8{25} = \tfrac{25}{50}-\tfrac1{50}-\tfrac{16}{50} = \tfrac{8}{50}=\tfrac4{25}$.

    Common Mistakes

    • Forgetting the constant term $A$ entirely (treating it as a standard proper partial fraction), which would make the setup inconsistent with the given equal-degree numerator/denominator.
    • Arithmetic slips evaluating $(0.5+2)(0.5-3)$ or converting fractions to a common denominator.

    ✓Final answer

    The correct option is (B) — $4/25$.

    ANSWER: B

11. 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

12. 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