Q.Integrate the function $\frac{x^2}{1-x^6}$

Showing the 12 most recent of 63 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int \frac{dx}{\sin x + \sin 2x} =$ (A) $\frac{1}{4}\log|1-\cos x| + \frac{1}{3}\log|1+\cos x| - \frac{2}{3}\log|1+\cos 2x| + c$ (B) $\frac{1}{3}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{1}{3}\log|1+\cos 2x| + c$ (C) $\frac{1}{6}\log|1-\cos x| + \frac{1}{2}\log|1+\cos x| - \frac{2}{3}\log|1+2\cos x| + c$ (D) $\frac{1}{6}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{2}{3}\log|1+2\cos x| + c$

   ›Reveal solutionSolution

   Factoring $\sin x+\sin2x=\sin x(1+2\cos x)$ and substituting $t=\cos x$ reduces this to a rational-function partial-fractions integral, giving $\frac16\log|1-\cos x|+\frac12\log|1+\cos x|-\frac23\log|1+2\cos x|+c$.

   Concept and Intuition

   Whenever an integral has $\sin x$ (an odd power effectively) times other cosine factors in the denominator, multiplying numerator and denominator by $\sin x$ turns $\sin^2x$ into $1-\cos^2x$, which lets us substitute $t=\cos x$ and reduce the whole problem to partial fractions of a rational function in $t$ — a completely mechanical final step.

   Step-by-Step Solution

   1. Factor the denominator: $\sin x+\sin2x = \sin x + 2\sin x\cos x = \sin x(1+2\cos x)$.
   2. So the integral is $\displaystyle\int\frac{dx}{\sin x(1+2\cos x)}$.
   3. Multiply top and bottom by $\sin x$: $\displaystyle\int\frac{\sin x\,dx}{\sin^2x(1+2\cos x)}=\int\frac{\sin x\,dx}{(1-\cos^2x)(1+2\cos x)}=\int\frac{\sin x\,dx}{(1-\cos x)(1+\cos x)(1+2\cos x)}$.
   4. Substitute $t=\cos x$, $dt=-\sin x\,dx$: integral $=-\displaystyle\int\frac{dt}{(1-t)(1+t)(1+2t)}$.
   5. Partial fractions: $\dfrac{1}{(1-t)(1+t)(1+2t)}=\dfrac{A}{1-t}+\dfrac{B}{1+t}+\dfrac{C}{1+2t}$. Evaluating at $t=1$: $A=\tfrac16$. At $t=-1$: $B=\tfrac12$. At $t=-\tfrac12$: $C=-\tfrac23$. …

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2. 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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3. 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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4. 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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5. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The absolute value of the difference of the coefficients of $x^4$ and $x^6$ in the expansion of $\dfrac{2x^2}{(x^2+1)(x^2+2)}$ is (A) $\dfrac{13}{4}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{9}{4}$ (D) 1

   ›Reveal solutionSolution

   Expanding the rational function as a power series (via partial fractions) in $u=x^2$ gives coefficients $-3/2$ (for $x^4$) and $7/4$ (for $x^6$); the absolute difference is $13/4$.

   Concept and Intuition

   The function is even in $x$ (a function of $x^2$ only), so its Maclaurin series has only even powers of $x$. It's cleanest to substitute $u=x^2$, resolve into partial fractions, and expand each simple fraction using the geometric series $\frac{1}{1+t}=1-t+t^2-t^3+\cdots$.

   Step-by-Step Solution

   1. Let $u=x^2$: $\dfrac{2x^2}{(x^2+1)(x^2+2)}=\dfrac{2u}{(u+1)(u+2)}$.
   2. Partial fractions: $\dfrac{2u}{(u+1)(u+2)}=\dfrac{A}{u+1}+\dfrac{B}{u+2}$. Solving, $2u=A(u+2)+B(u+1)$; at $u=-1$: $-2=A\Rightarrow A=-2$; at $u=-2$: $-4=-B\Rightarrow B=4$.
   3. So the function $=\dfrac{-2}{1+u}+\dfrac{4}{2+u} = -2(1+u)^{-1} + 2(1+u/2)^{-1}$.
   4. Expand: $-2(1-u+u^2-u^3+u^4-\cdots)$ and $2(1-\tfrac u2+\tfrac{u^2}4-\tfrac{u^3}8+\cdots)$.
   5. Coefficient of $u^2$ (this is the $x^4$ term): $-2(1)+2\left(\tfrac14\right) = -2+\tfrac12=-\tfrac32$. …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{2x+5}{(x-1)(x+1)(x+4)(x+6)}\, dx = \dfrac{1}{10}\log\left(\dfrac{f(x)}{g(x)}\right) + c$ and $\dfrac{f(-2)}{g(-2)} = 6$, then $\dfrac{f(10)}{g(10)} =$ (A) $\dfrac{72}{77}$ (B) $\dfrac{144}{75}$ (C) $\dfrac{55}{63}$ (D) $\dfrac{70}{59}$

   ›Reveal solutionSolution

   Grouping the quartic denominator via $u=x^2+5x$ turns it into $(u-6)(u+4)$, whose partial fractions give $f(x)=(x-1)(x+6)$ and $g(x)=(x+1)(x+4)$; evaluating at $x=10$ gives $144/154=72/77$.

   Concept and Intuition

   $(x-1)(x+6)=x^2+5x-6$ and $(x+1)(x+4)=x^2+5x+4$ share the same quadratic core $x^2+5x$. Substituting $u=x^2+5x$ (whose derivative $2x+5$ is exactly the numerator) collapses the quartic denominator into a simple product $(u-6)(u+4)$, reducing the problem to a standard $\int \frac{du}{(u-a)(u-b)}$ form.

   Step-by-Step Solution

   1. Let $u=x^2+5x \Rightarrow du=(2x+5)dx$. Then

   $(x-1)(x+1)(x+4)(x+6)=(u-6)(u+4)$

   since $(x-1)(x+6)=x^2+5x-6=u-6$ and $(x+1)(x+4)=x^2+5x+4=u+4$.

   2. The integral becomes

   $\int\frac{du}{(u-6)(u+4)}=\frac{1}{10}\int\left(\frac{1}{u-6}-\frac{1}{u+4}\right)du=\frac{1}{10}\log\left|\frac{u-6}{u+4}\right|+c$

   3. So $\dfrac{f(x)}{g(x)}=\dfrac{u-6}{u+4}=\dfrac{(x-1)(x+6)}{(x+1)(x+4)}$. …

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7. 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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8. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If we resolve the rational fraction $\dfrac{1}{(1-2x)^2(1-3x)}$ in to partial fractions of the form $\dfrac{A}{1-3x}+\dfrac{B}{1-2x}+\dfrac{C}{(1-2x)^2}$, then what is the $\min\{A,B,C\} =$ (A) 1 (B) 9 (C) -2 (D) -6

   ›Reveal solutionSolution

   Standard partial-fraction decomposition via strategic substitution — the minimum of $A,B,C$ is (D) -6.

   Concept and Intuition

   For a partial-fraction decomposition with a repeated linear factor, plugging in the value of $x$ that zeroes each denominator isolates that factor's coefficient directly (the Heaviside cover-up trick), and one extra convenient value (like $x=0$) resolves any remaining unknown.

   Step-by-Step Solution

   1. Write $\dfrac{1}{(1-2x)^2(1-3x)} = \dfrac{A}{1-3x}+\dfrac{B}{1-2x}+\dfrac{C}{(1-2x)^2}$.
   2. Multiply both sides by $(1-2x)^2(1-3x)$: $1 = A(1-2x)^2 + B(1-2x)(1-3x) + C(1-3x)$.
   3. Set $x=\tfrac13$ (zeroes the $B,C$ terms): $1 = A\left(\tfrac13\right)^2 = \tfrac{A}{9} \Rightarrow A=9$. …

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9. 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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10. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $\dfrac{x^4}{(x-1)(x-2)}=f(x)+\dfrac{A}{x-1}+\dfrac{B}{x-2}$, then $f(-2)+A+B=$ (A) 32 (B) 28 (C) 22 (D) 20

    ›Reveal solutionSolution

    This tests polynomial long division combined with partial fractions — split $\dfrac{x^4}{(x-1)(x-2)}$ into a polynomial part $f(x)$ plus proper fractions. Answer: $f(-2)+A+B=20$.

    Concept and Intuition

    When the numerator's degree ($4$) is greater than or equal to the denominator's degree ($2$), a rational function isn't purely a sum of partial fractions — you first must divide out a polynomial quotient $f(x)$, leaving a proper-fraction remainder that partial-fractions cleanly. Here $f(x)$ is exactly that quotient (degree $4-2=2$).

    Step-by-Step Solution

    1. Divide $x^4$ by $x^2-3x+2$ (long division): $x^4 = (x^2-3x+2)(x^2+3x+7) + (15x-14)$ Check: $(x^2-3x+2)(x^2+3x+7) = x^4 - 15x+14$, so adding $15x-14$ recovers $x^4$. ✓
    2. So $f(x)=x^2+3x+7$, and the remainder gives $\dfrac{15x-14}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$.
    3. Clear denominators: $15x-14=A(x-2)+B(x-1)$.
    4. Put $x=1$: $15-14=A(-1)\Rightarrow 1=-A\Rightarrow A=-1$. …

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11. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.The coefficient of $x^3$ in the power series expansion of $\dfrac{x}{x^2-x-2}$ is (A) $-\dfrac{1}{8}$ (B) $-\dfrac{3}{8}$ (C) $\dfrac{3}{8}$ (D) $\dfrac{1}{8}$

    ›Reveal solutionSolution

    Partial-fraction the rational function, expand each simple fraction as a geometric series in $x$, and read off the coefficient of $x^3$ — it comes out to $-3/8$.

    Concept and Intuition

    A rational function whose denominator factors into distinct linear factors can be split into partial fractions, each of which is a simple geometric-series generating function ($\frac{1}{1-x}$-type), making it straightforward to extract any coefficient of the power series expansion around $x=0$.

    Step-by-Step Solution

    1. Factor the denominator: $x^2-x-2 = (x-2)(x+1)$.
    2. Set up partial fractions: $\dfrac{x}{(x-2)(x+1)} = \dfrac{A}{x-2}+\dfrac{B}{x+1}$, so $x = A(x+1)+B(x-2)$.
    3. At $x=2$: $2 = 3A \Rightarrow A = \dfrac23$. At $x=-1$: $-1 = -3B \Rightarrow B=\dfrac13$.
    4. So the function is $\dfrac{2/3}{x-2} + \dfrac{1/3}{x+1}$.
    5. Expand the first term (valid for $|x|<2$): $\dfrac{2/3}{x-2} = -\dfrac13\cdot\dfrac{1}{1-x/2} = -\dfrac13\sum_{n\ge0}\left(\dfrac{x}{2}\right)^n$. Coefficient of $x^3$: $-\dfrac13\cdot\dfrac{1}{2^3}=-\dfrac{1}{24}$. …

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12. 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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