Q.Integrate the following function: $\frac{3x + 5}{x^3 - x^2 - x + 1}$

Showing the 12 most recent of 63 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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 $\dfrac{3x-4}{(x+4)(x-1)} = \dfrac{16}{5(x+4)} - \dfrac{1}{5(x-1)}$. …

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7. 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: $-(-\tfrac12)+1-2(\tfrac12)+(-\tfrac12) = \tfrac12+1-1-\tfrac12=0$ ✓.) …

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8. 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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9. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\dfrac{3x+1}{(x-1)(x^2+2)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+2}$, then $5(A-B)=$ (A) $A+C$ (B) $8C$ (C) $C+8$ (D) $\dfrac{C}{8}$

   ›Reveal solutionSolution

   This tests partial-fraction decomposition and coefficient comparison. Answer: $5(A-B)=8C$.

   Concept and Intuition

   To find $A,B,C$ in a partial fraction decomposition, clear denominators to get a polynomial identity, then either substitute convenient values of $x$ (like the root of the linear factor) or compare coefficients of like powers of $x$.

   Step-by-Step Solution

   1. Clear denominators: $3x+1 = A(x^2+2)+(Bx+C)(x-1)$.
   2. Substitute $x=1$ (kills the $(Bx+C)(x-1)$ term): $3(1)+1 = A(1+2) \Rightarrow 4=3A \Rightarrow A=\dfrac43$.
   3. Expand the right side: $Ax^2+2A+Bx^2-Bx+Cx-C = (A+B)x^2+(C-B)x+(2A-C)$.
   4. Compare coefficient of $x^2$ (LHS has none): $A+B=0\Rightarrow B=-A=-\dfrac43$.
   5. Compare coefficient of $x^1$: $C-B=3\Rightarrow C=3+B=3-\dfrac43=\dfrac53$.
   6. Check constant term: $2A-C = \dfrac83-\dfrac53=1$ ✓, consistent. …

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10. 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 $A_2=1,A_4=-1$: $0=-A_1+A_3 \Rightarrow A_1=A_3$. …

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11. 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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12. 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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