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

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 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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3. 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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4. 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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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 …

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6. 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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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 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $\dfrac{3x^2+x+2}{(3x^2+x+4)(3x^2+x+1)} = \dfrac{Ax+B}{3x^2+x+4} + \dfrac{Cx+D}{3x^2+x+1}$, then $(A+B) + (C+D) =$ (A) $\dfrac{1}{3}$ (B) $\dfrac{2}{3}$ (C) $1$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Substituting $u=3x^2+x$ collapses the problem to an ordinary constant partial fraction in $u$, forcing $A=C=0$ and giving $(A+B)+(C+D)=1$.

   Concept and Intuition

   When a rational expression's numerator and both denominator factors are built from the same quadratic block $3x^2+x$ shifted by constants, it's really a partial-fraction problem in the single variable $u=3x^2+x$, not in $x$ directly. Recognizing this shortcut avoids a messy 4-unknown system in $x$.

   Step-by-Step Solution

   1. Let $u=3x^2+x$. The equation becomes $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{Ax+B}{u+4}+\dfrac{Cx+D}{u+1}$.
   2. Do ordinary partial fractions in $u$: $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{P}{u+4}+\dfrac{Q}{u+1}$.
   3. Cover-up at $u=-4$: $P = \dfrac{-4+2}{-4+1} = \dfrac{-2}{-3}=\dfrac23$. At $u=-1$: $Q=\dfrac{-1+2}{-1+4}=\dfrac{1}{3}$.
   4. So $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{2/3}{u+4}+\dfrac{1/3}{u+1}$, which is entirely $x$-independent in its numerators. …

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9. 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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10. 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: $\displaystyle\int\left(1+\frac1x-\frac1{x^2}-\frac{2}{x+1}\right)dx=x+\log|x|+\frac1x-2\log|x+1|+c=\left(x+\frac1x\right)+\log\left|\frac{x}{(x+1)^2}\right|+c$. …

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

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12. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.If $\dfrac{x^2}{(x^2+2)(x^4-1)}=\dfrac{A}{x^2-1}+\dfrac{B}{x^2+1}+\dfrac{C}{x^2+2}$, then $A+B-C=$ (A) $0$ (B) $\dfrac{4}{3}$ (C) $\dfrac{3}{4}$ (D) $2$

    ›Reveal solutionSolution

    A partial-fractions problem in disguise (substitute $y=x^2$); solving gives $A+B-C=\frac{4}{3}$.

    Concept and Intuition

    Since $x^4-1=(x^2-1)(x^2+1)$, the whole expression is a rational function purely in $y=x^2$. Substituting $y=x^2$ converts it into an ordinary partial-fractions decomposition with three distinct linear factors $(y-1),(y+1),(y+2)$, solvable by the cover-up (Heaviside) method.

    Step-by-Step Solution

    1. Let $y=x^2$. The equation becomes $\dfrac{y}{(y+2)(y-1)(y+1)}=\dfrac{A}{y-1}+\dfrac{B}{y+1}+\dfrac{C}{y+2}$.
    2. Clear denominators: $y=A(y+1)(y+2)+B(y-1)(y+2)+C(y-1)(y+1)$.
    3. At $y=1$: $1=A(2)(3)=6A\Rightarrow A=\dfrac16$.
    4. At $y=-1$: $-1=B(-2)(1)=-2B\Rightarrow B=\dfrac12$.
    5. At $y=-2$: $-2=C(-3)(-1)=3C\Rightarrow C=-\dfrac23$. …

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