Q.Integrate the function $\frac{5x}{(x+1)(x^2+9)}$

Showing the 12 most recent of 63 on this concept.

1. 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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2. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If $\dfrac{1}{(3-5x)(2+3x)}=\dfrac{A}{3-5x}+\dfrac{B}{2+3x}$ then $A+B=$ (A) $\dfrac{7}{19}$ (B) $\dfrac{8}{19}$ (C) $\dfrac{9}{19}$ (D) $\dfrac{10}{19}$

   ›Reveal solutionSolution

   Standard partial fractions: clearing denominators and plugging in the roots of each factor gives $A=5/19$, $B=3/19$, so $A+B=8/19$.

   Concept and Intuition

   For a proper rational function with distinct linear factors in the denominator, each constant is found by substituting the value of $x$ that zeroes out the OTHER factor.

   Step-by-Step Solution

   1. Write $\dfrac{1}{(3-5x)(2+3x)}=\dfrac{A}{3-5x}+\dfrac{B}{2+3x}$.
   2. Multiply through: $1=A(2+3x)+B(3-5x)$.
   3. Set $3-5x=0 \Rightarrow x=3/5$: $1=A(2+3\cdot3/5)=A(2+9/5)=A\cdot\dfrac{19}{5}\Rightarrow A=\dfrac{5}{19}$.
   4. Set $2+3x=0\Rightarrow x=-2/3$: $1=B(3-5\cdot(-2/3))=B(3+10/3)=B\cdot\dfrac{19}{3}\Rightarrow B=\dfrac{3}{19}$. …

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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). …

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4. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If $\dfrac{x^2+3}{x^4+2x^2+9} = \dfrac{Ax+B}{x^2+ax+b} + \dfrac{Cx+D}{x^2+cx+b}$ then $aA + bB + cC + D =$ (A) $1$ (B) $0$ (C) $-1$ (D) $2$

   ›Reveal solutionSolution

   Factor the quartic denominator into two quadratics, split the fraction, solve for the constants by matching coefficients, and evaluate the requested combination. Answer: $2$.

   Concept and Intuition

   The quartic $x^4+2x^2+9$ has the classic Sophie-Germain-like factorization $(x^2+px+q)(x^2-px+q)=x^4+(2q-p^2)x^2+q^2$; matching $q^2=9,\ 2q-p^2=2$ gives $q=3,\ p=2$. Once the denominators are known, comparing coefficients of a polynomial identity pins down $A,B,C,D$ uniquely.

   Step-by-Step Solution

   1. Factor: $x^4+2x^2+9=(x^2+2x+3)(x^2-2x+3)$, so $a=2,\ b=3,\ c=-2$ (matching $(x^2+ax+b)(x^2+cx+b)$).
   2. Write $\dfrac{x^2+3}{(x^2+2x+3)(x^2-2x+3)}=\dfrac{Ax+B}{x^2+2x+3}+\dfrac{Cx+D}{x^2-2x+3}$.
   3. Combine and match numerators: $(Ax+B)(x^2-2x+3)+(Cx+D)(x^2+2x+3)=x^2+3$. Expanding and comparing coefficients of $x^3,x^2,x^1,x^0$ gives $A+C=0$, $D=B$, and solving the system yields $A=0,\ C=0,\ B=D=\dfrac12$. …

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

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

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9. 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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10. 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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11. 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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12. 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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