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

Showing the 12 most recent of 63 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $\dfrac{2x^3 - 7x^2 - 2x + 20}{x^2 - 5x + 6} = Ax + B + \dfrac{C}{x-2} + \dfrac{D}{x-3}$, then $A + B + C + D =$ (A) $2$ (B) $3$ (C) $5$ (D) $6$

   ›Reveal solutionSolution

   This tests polynomial long division followed by partial-fraction decomposition. The final sum of all four constants is $6$.

   Concept and Intuition

   When a rational function's numerator degree is $\geq$ the denominator degree, first perform polynomial division to extract the linear (or polynomial) part $Ax+B$, leaving a proper fraction that can then be split into partial fractions over the denominator's linear factors.

   Step-by-Step Solution

   1. Divide $2x^3-7x^2-2x+20$ by $x^2-5x+6$:
      • $2x^3 \div x^2 = 2x$; $2x(x^2-5x+6) = 2x^3-10x^2+12x$; subtracting leaves $3x^2-14x+20$.
      • $3x^2 \div x^2 = 3$; $3(x^2-5x+6)=3x^2-15x+18$; subtracting leaves $x+2$.
      • So quotient $=2x+3$, remainder $=x+2$, giving $A=2,\ B=3$.
   2. Factor the denominator: $x^2-5x+6=(x-2)(x-3)$. …

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2. 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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3. 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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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $\dfrac{f(x)}{(x-1)(x-2)} = \dfrac{2}{x-2} - \dfrac{1}{x-1}$ and $f(x) + \dfrac{xf(x)}{(x-1)(x-2)} = g(x) + \dfrac{A}{x-2} + \dfrac{B}{x-1}$, then $g(A + B) =$ (A) $6$ (B) $5$ (C) $4$ (D) $8$

   ›Reveal solutionSolution

   This is a partial-fractions identification problem: first recover $f(x)$ from the given fraction identity, then re-expand a new expression in partial fractions and match coefficients. The answer is (C).

   Concept and Intuition

   A rational function equated to a sum of simpler fractions is a partial-fraction decomposition. If we're told the decomposition, we can recover the original numerator by combining the right side over the common denominator. The same idea, run in reverse, lets us decompose a new rational expression once we know $f(x)$.

   Step-by-Step Solution

   1. Combine the RHS: $\dfrac{2}{x-2}-\dfrac{1}{x-1} = \dfrac{2(x-1)-(x-2)}{(x-1)(x-2)} = \dfrac{2x-2-x+2}{(x-1)(x-2)} = \dfrac{x}{(x-1)(x-2)}$.
   2. Comparing with $\dfrac{f(x)}{(x-1)(x-2)}$, we get $f(x) = x$.
   3. Now compute $f(x) + \dfrac{xf(x)}{(x-1)(x-2)} = x + \dfrac{x\cdot x}{(x-1)(x-2)} = x + \dfrac{x^2}{(x-1)(x-2)}$.
   4. Since $x^2$ and $(x-1)(x-2)=x^2-3x+2$ have the same degree, do polynomial division first: $\dfrac{x^2}{x^2-3x+2} = 1 + \dfrac{3x-2}{(x-1)(x-2)}$. …

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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 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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7. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $\dfrac{x^4}{x^4+3x^2+2} = A + \dfrac{Bx+C}{x^2+p} + \dfrac{Dx+E}{x^2+q}$ and $p < q$, then $\dfrac{A-B+C+D-E}{p+q} =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Factoring the denominator gives $p=1,q=2$; since the fraction is improper (degree 4 over degree 4), first split off the constant term, then use symmetry to kill the odd-power numerators, then match coefficients to get the constants — the final ratio works out to 2.

   Concept and Intuition

   When degree(numerator) = degree(denominator), a partial fraction decomposition must start with polynomial division to extract a constant (here $A$) before the proper-fraction part can be split further. Because the remaining fraction, $\dfrac{-(3x^2+2)}{(x^2+1)(x^2+2)}$, is an even function of $x$ (only even powers appear), its partial fraction decomposition over the two irreducible quadratics $x^2+1$ and $x^2+2$ cannot contain any odd-power ($x^1$) numerator terms — so $B=D=0$ automatically, without doing any algebra for them.

   Step-by-Step Solution

   1. Factor: $x^4+3x^2+2 = (x^2+1)(x^2+2)$, so comparing with $x^2+p$ and $x^2+q$ (with $p<q$): $p=1,\ q=2$.
   2. Since the given fraction is improper (both numerator and denominator degree 4), perform division: $\dfrac{x^4}{x^4+3x^2+2} = 1 - \dfrac{3x^2+2}{(x^2+1)(x^2+2)}$. So $A=1$.
   3. Write $-\dfrac{3x^2+2}{(x^2+1)(x^2+2)} = \dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{x^2+2}$. Since the left side is even in $x$, no odd powers can appear on the right, so $B=D=0$. …

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8. 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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9. 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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10. 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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11. 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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12. 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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