Q.$\int \dfrac{2x-1}{(x-1)(x+2)(x-3)}\,dx$

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 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): $\left(\tfrac12\right)^3 = B\left(\tfrac12+2\right)\left(\tfrac12-3\right) = B(2.5)(-2.5)=-6.25B$. $\tfrac18 = -\tfrac{25}{4}B \Rightarrow B = -\tfrac{1}{50}$. …

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3. 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 $A$. …

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4. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.If the equivalent partial fraction of $\dfrac{x^3}{(2x-1)(x+2)(x-3)}$ is given by $A + \dfrac{B}{2x-1}+\dfrac{C}{x+2}+\dfrac{D}{x-3}$, then the value of C is (A) $1/2$ (B) $-1/50$ (C) $-8/25$ (D) $27/25$

   ›Reveal solutionSolution

   This tests the cover-up (Heaviside) method for finding a specific partial-fraction coefficient by substituting the root of that factor. Answer: $C=-8/25$.

   Concept and Intuition

   When a proper (or improper, with a polynomial part) rational function is decomposed into partial fractions with distinct linear factors in the denominator, the coefficient attached to a factor like $\dfrac{C}{x+2}$ can be found instantly by the 'cover-up rule': multiply the whole equation by $(x+2)$, which cancels that factor from the left side and leaves only $C$ (plus terms that vanish) on the right when we then substitute $x=-2$ (the root of that factor). This avoids fully expanding and comparing coefficients.

   Step-by-Step Solution

   1. Write $\dfrac{x^3}{(2x-1)(x+2)(x-3)} = A + \dfrac{B}{2x-1}+\dfrac{C}{x+2}+\dfrac{D}{x-3}$.
   2. Multiply both sides by $(x+2)$: $\dfrac{x^3}{(2x-1)(x-3)} = A(x+2) + \dfrac{B(x+2)}{2x-1} + C + \dfrac{D(x+2)}{x-3}$.
   3. Set $x=-2$: on the right, every term except $C$ has a factor $(x+2)=0$, so they vanish, leaving just $C$.
   4. On the left: $\dfrac{(-2)^3}{(2(-2)-1)((-2)-3)} = \dfrac{-8}{(-4-1)(-5)} = \dfrac{-8}{(-5)(-5)} = \dfrac{-8}{25}$. …

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5. 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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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 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 $x=-2$: $5=0+0+C(-4)(5)\Rightarrow -20C=5\Rightarrow C=-\tfrac14$. …

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8. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \Rightarrow 7A+2B+C =$ (A) $8$ (B) $9$ (C) $10$ (D) $11$

   ›Reveal solutionSolution

   Clear denominators in the partial fraction decomposition, use $x=1$ to find $A$ quickly, then match coefficients for $B$ and $C$.

   Concept and Intuition

   Since $x^3-1=(x-1)(x^2+x+1)$, the partial fraction form given is standard. Plugging in the root of the linear factor ($x=1$) isolates $A$ immediately; matching remaining coefficients (or plugging in more values) gives $B,C$.

   Step-by-Step Solution

   1. Multiply both sides by $x^3-1=(x-1)(x^2+x+1)$: $2x^2+1 = A(x^2+x+1) + (Bx+C)(x-1)$.
   2. Set $x=1$: $2(1)+1 = A(1+1+1) + 0 \Rightarrow 3 = 3A \Rightarrow A=1$.
   3. Expand the right side: $A x^2+Ax+A + Bx^2-Bx+Cx-C = (A+B)x^2+(A-B+C)x+(A-C)$.
   4. Match $x^2$ coefficient: $A+B=2 \Rightarrow B = 2-1=1$.
   5. Match constant term: $A - C = 1 \Rightarrow C = A-1 = 0$. …

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9. 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 $C=-2E=-\dfrac12$. …

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10. 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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11. 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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12. 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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