Q.Evaluate the definite integral: $\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}$ equals (A) $\frac{\pi}{3}$ (B) $\frac{2\pi}{3}$ (C) $\frac{\pi}{6}$ (D) $\frac{\pi}{12}$

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\displaystyle\int_0^1 \dfrac{x^4+1}{x^6+1}\, dx =$ (A) $\dfrac{\pi}{3}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Splitting $(x^4+1)/(x^6+1)$ into a partial-fraction sum over $(x^2+1)$ and $(x^4-x^2+1)$, each piece integrates to an arctan expression, and the two pieces combine to exactly $\pi/3$.

   Concept and Intuition

   $x^6+1$ factors as $(x^2+1)(x^4-x^2+1)$. Writing the numerator as a combination that matches this factorization turns the integral into two standard pieces: a plain arctan, and a "divide by $x^2$, substitute $t=x-1/x$" trick that's classic for symmetric quartics.

   Step-by-Step Solution

   1. Factor: $x^6+1=(x^2+1)(x^4-x^2+1)$.
   2. Find constants with $x^4+1=a(x^4-x^2+1)+(bx^2+c)(x^2+1)$. Matching coefficients gives $a=\tfrac23,\ b=\tfrac13,\ c=\tfrac13$, so

   $\dfrac{x^4+1}{x^6+1}=\dfrac{2/3}{x^2+1}+\dfrac{(1/3)(x^2+1)}{x^4-x^2+1}.$

   3. First term: $\displaystyle\dfrac23\int_0^1\dfrac{dx}{x^2+1}=\dfrac23\cdot\dfrac\pi4=\dfrac\pi6$.
   4. Second term: $K=\displaystyle\int_0^1\dfrac{x^2+1}{x^4-x^2+1}dx$. Divide numerator/denominator by $x^2$:

   $K=\int\dfrac{1+1/x^2}{x^2-1+1/x^2}dx.$

   Let $t=x-\dfrac1x$, so $dt=\left(1+\dfrac1{x^2}\right)dx$ and $x^2+\dfrac1{x^2}-1=t^2+1$.

   $K=\int\dfrac{dt}{t^2+1}=\tan^{-1}t=\tan^{-1}\!\left(x-\dfrac1x\right).$ …

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2. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/4} \dfrac{\cos^2 x}{\cos^2 x + 4\sin^2 x}dx =$ (A) $\dfrac{\pi}{4}+\dfrac{2}{3}\tan^{-1}2$ (B) $-\dfrac{\pi}{3}-\dfrac{2}{3}\tan^{-1}3$ (C) $-\dfrac{\pi}{12}+\dfrac{2}{3}\tan^{-1}2$ (D) $\dfrac{\pi}{6}-\dfrac{2}{3}\tan^{-1}4$

   ›Reveal solutionSolution

   Dividing through by $\cos^2x$ turns this into a rational function of $\tan x$, which splits by partial fractions into two standard arctangent integrals; the answer is $-\pi/12+\tfrac23\tan^{-1}2$.

   Concept and Intuition

   Integrals of the form $\int \dfrac{\cos^2x}{\cos^2x+k\sin^2x}\,dx$ become rational in $t=\tan x$ after dividing by $\cos^2x$, since $\dfrac{1}{\cos^2x+k\sin^2x}=\dfrac{\sec^2x}{1+k\tan^2x}$.

   Step-by-Step Solution

   1. $\dfrac{\cos^2x}{\cos^2x+4\sin^2x}=\dfrac{1}{1+4\tan^2x}$ (divide top & bottom by $\cos^2x$).
   2. Let $t=\tan x$, $dt=\sec^2x\,dx$; limits $x:0\to\pi/4 \Rightarrow t:0\to1$.
   3. $\displaystyle\int_0^{\pi/4}\dfrac{dx}{1+4\tan^2x}=\int_0^1\dfrac{dt}{(1+4t^2)(1+t^2)}$ (bringing in the extra $1/(1+t^2)$ from $dx=dt/(1+t^2)$).
   4. Partial fractions: $\dfrac{1}{(1+t^2)(1+4t^2)}=\dfrac{-1/3}{1+t^2}+\dfrac{4/3}{1+4t^2}$.
   5. $\displaystyle\int_0^1\dfrac{-1/3}{1+t^2}dt=-\dfrac13\tan^{-1}(1)=-\dfrac{\pi}{12}$. …

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3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/4} \dfrac{\cos^2 x}{\cos^2 x + 4\sin^2 x}dx=$ (A) $\dfrac{\pi}{2} - \dfrac{1}{3}\mathrm{Tan}^{-1}2$ (B) $-\dfrac{\pi}{4} - \dfrac{4}{3}\mathrm{Tan}^{-1}2$ (C) $\dfrac{\pi}{6} + \dfrac{2}{3}\mathrm{Tan}^{-1}2$ (D) $-\dfrac{\pi}{12} + \dfrac{2}{3}\mathrm{Tan}^{-1}2$

   ›Reveal solutionSolution

   Dividing by $\cos^2x$ and substituting $t=\tan x$ turns this into a rational-function integral, giving $-\frac{\pi}{12}+\frac23\mathrm{Tan}^{-1}2$.

   Concept and Intuition

   When an integrand is a ratio of quadratics in $\sin x,\cos x$, dividing through by $\cos^2x$ and substituting $t=\tan x$ converts it into a rational function of $t$, which is handled by ordinary partial fractions.

   Step-by-Step Solution

   1. Divide numerator and denominator by $\cos^2x$: $\dfrac{\cos^2x}{\cos^2x+4\sin^2x}=\dfrac1{1+4\tan^2x}$.
   2. Let $t=\tan x$, $dt=\sec^2x\,dx=(1+t^2)\,dx$; limits $x:0\to\pi/4$ become $t:0\to1$.
   3. $\displaystyle I=\int_0^1\frac1{1+4t^2}\cdot\frac{dt}{1+t^2}=\int_0^1\frac{dt}{(1+4t^2)(1+t^2)}$.
   4. Partial fractions: $\dfrac1{(1+4t^2)(1+t^2)}=-\dfrac13\cdot\dfrac1{1+t^2}+\dfrac43\cdot\dfrac1{1+4t^2}$ (solve $1=A(1+4t^2)+B(1+t^2)$ giving $A=-\tfrac13,B=\tfrac43$). …

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4. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int_1^2 \dfrac{x^4-1}{x^6-1}\,dx=$ (A) $\dfrac{1}{\sqrt{3}}\tan^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$ (B) $\dfrac{121}{6}$ (C) $\sqrt{2}-1$ (D) $\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{2}{\sqrt{3}}\right)$

   ›Reveal solutionSolution

   Factoring and a $u=x-1/x$ substitution reduce the integral to $\int du/(u^2+3)$, giving $\dfrac1{\sqrt3}\tan^{-1}(\sqrt3/2)$.

   Concept and Intuition

   When a rational function has the same even symmetry structure top and bottom (here both are built from $x^2$-blocks), dividing through by $x^2$ and substituting $u=x-1/x$ (whose derivative is $1+1/x^2$, matching the numerator) is the standard trick to collapse it into an arctangent integral.

   Step-by-Step Solution

   1. Factor: $x^4-1=(x^2-1)(x^2+1)$ and $x^6-1=(x^2-1)(x^4+x^2+1)$, so $\dfrac{x^4-1}{x^6-1}=\dfrac{x^2+1}{x^4+x^2+1}$ (the $x^2-1$ cancels, valid since $x\in[1,2]$ avoids the removable singularity at $x=1$ in the limit sense).
   2. Divide numerator and denominator by $x^2$: $\dfrac{1+\frac1{x^2}}{x^2+1+\frac1{x^2}}$.
   3. Let $u=x-\dfrac1x$, so $du=\left(1+\dfrac1{x^2}\right)dx$ — exactly the numerator.
   4. Note $u^2=x^2-2+\dfrac1{x^2}$, so $x^2+\dfrac1{x^2}=u^2+2$, making the denominator $u^2+2+1=u^2+3$.
   5. The integral becomes $\displaystyle\int \frac{du}{u^2+3}=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{u}{\sqrt3}\right)+C$.
   6. Limits: at $x=1$, $u=1-1=0$; at $x=2$, $u=2-\dfrac12=\dfrac32$. …

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5. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.$\displaystyle\int_{-1}^{2} \left[\tan^{-1}\left(\dfrac{x}{x^2+1}\right) + \tan^{-1}\left(\dfrac{x^2+1}{x}\right)\right] dx =$ (A) $\dfrac{\pi}{4}$ (B) $\dfrac{3\pi}{4}$ (C) $\pi/4$ (D) $\pi/2$

   ›Reveal solutionSolution

   The integrand simplifies to a step function ($-\pi/2$ for $x<0$, $+\pi/2$ for $x>0$) via the identity $\tan^{-1}a+\tan^{-1}(1/a)=\pm\pi/2$, giving $\pi/2$ after integrating over $[-1,2]$.

   Concept and Intuition

   $\tan^{-1}a + \tan^{-1}\dfrac1a$ is a classic identity that is not a fixed constant — it depends on the sign of $a$: it equals $\pi/2$ when $a>0$ and $-\pi/2$ when $a<0$. Recognizing that the two terms in the integrand are exactly reciprocal arguments of each other turns a scary-looking integral into integrating a simple piecewise constant.

   Step-by-Step Solution

   1. Let $a(x)=\dfrac{x}{x^2+1}$. Then the second term's argument is $\dfrac{x^2+1}{x}=\dfrac{1}{a(x)}$.
   2. Since $x^2+1>0$ always, the sign of $a(x)$ matches the sign of $x$.
   3. For $x>0$: $a>0\Rightarrow \tan^{-1}a+\tan^{-1}(1/a)=\dfrac\pi2$.
   4. For $x<0$: $a<0\Rightarrow \tan^{-1}a+\tan^{-1}(1/a)=-\dfrac\pi2$.
   5. (At $x=0$ the second term is technically undefined but it's a single point, measure zero, doesn't affect the integral.) …

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6. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{x^4+1}{x^6+1}\,dx =$ (A) $\mathrm{Tan}^{-1}x - \mathrm{Tan}^{-1}x^3 + c$ (B) $\mathrm{Tan}^{-1}x - \dfrac{1}{3}\mathrm{Tan}^{-1}x^3 + c$ (C) $\mathrm{Tan}^{-1}x + \mathrm{Tan}^{-1}x^3 + c$ (D) $\mathrm{Tan}^{-1}x + \dfrac{1}{3}\mathrm{Tan}^{-1}x^3 + c$

   ›Reveal solutionSolution

   Splitting the numerator cleverly turns the integral into $\int\frac{1}{x^2+1}dx + \int\frac{x^2}{x^6+1}dx$, giving $\mathrm{Tan}^{-1}x+\tfrac13\mathrm{Tan}^{-1}x^3+c$.

   Concept and Intuition

   $x^6+1$ factors as $(x^2+1)(x^4-x^2+1)$. Writing the numerator $x^4+1$ as $(x^4-x^2+1)+x^2$ lets one term cancel the second factor of the denominator exactly.

   Step-by-Step Solution

   1. $x^6+1=(x^2+1)(x^4-x^2+1)$.
   2. $x^4+1 = (x^4-x^2+1)+x^2$.
   3. So $\dfrac{x^4+1}{x^6+1} = \dfrac{x^4-x^2+1}{(x^2+1)(x^4-x^2+1)} + \dfrac{x^2}{x^6+1} = \dfrac{1}{x^2+1}+\dfrac{x^2}{x^6+1}$.
   4. $\int\dfrac{dx}{x^2+1}=\mathrm{Tan}^{-1}x$.
   5. For $\int\dfrac{x^2}{x^6+1}dx$, let $w=x^3,\ dw=3x^2dx$: $=\dfrac13\int\dfrac{dw}{w^2+1}=\dfrac13\mathrm{Tan}^{-1}(x^3)$. …

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7. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.If $\displaystyle\int_0^a \dfrac{dx}{4+x^2} = \dfrac{\pi}{8}$, then the value of $a =$ _______ (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   This tests the standard integral $\int \frac{dx}{a^2+x^2} = \frac1a\tan^{-1}(x/a)$ applied to a definite integral, then solving for the upper limit. Answer: $a=2$.

   Concept and Intuition

   Recognize the integrand as the standard $\dfrac{1}{k^2+x^2}$ form with $k=2$, whose antiderivative is a scaled arctangent; evaluating between the limits and setting equal to the given value determines the unknown upper limit.

   Step-by-Step Solution

   1. $\displaystyle\int \dfrac{dx}{4+x^2} = \dfrac12\mathrm{Tan}^{-1}\left(\dfrac x2\right)+C$.
   2. Evaluate from $0$ to $a$: $\dfrac12\mathrm{Tan}^{-1}\left(\dfrac a2\right) - \dfrac12\mathrm{Tan}^{-1}(0) = \dfrac12\mathrm{Tan}^{-1}\left(\dfrac a2\right)$. …

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8. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$\displaystyle\int\dfrac{1}{(1+x^2)\sqrt{x^2+2}}\,dx=$ (A) $-\text{Tan}^{-1}\dfrac{\sqrt{x^2+2}}{|x|}+c$ (B) $-\text{Tan}^{-1}\sqrt{x^2+2}+c$ (C) $\text{Tan}^{-1}\sqrt{\dfrac{x^2+1}{x^2+2}}+c$ (D) $-\text{Tan}^{-1}\sqrt{\dfrac{x^2+2}{x^2+1}}+c$

   ›Reveal solutionSolution

   The substitution $x=1/t$ (or equivalently dividing by $x$ inside the root) reduces this to a standard arctan integral; the antiderivative is $-\text{Tan}^{-1}\dfrac{\sqrt{x^2+2}}{|x|}+c$.

   Concept and Intuition

   Integrals of the form $\displaystyle\int\dfrac{dx}{(1+x^2)\sqrt{x^2+a^2}}$ are classically attacked by the reciprocal substitution $x=\dfrac1t$, which turns the awkward product of $(1+x^2)$ and a square root into a single, simpler square root in $t$ — collapsing to a standard $\int\dfrac{dp}{1+p^2}$ form after a further substitution.

   Step-by-Step Solution

   1. Let $x=\dfrac1t$ (for $x>0$), so $dx=-\dfrac{dt}{t^2}$.
   2. $1+x^2=1+\dfrac{1}{t^2}=\dfrac{t^2+1}{t^2}$, and $\sqrt{x^2+2}=\sqrt{\dfrac{1}{t^2}+2}=\dfrac{\sqrt{1+2t^2}}{|t|}$.
   3. The integral becomes $\displaystyle\int\dfrac{-dt/t^2}{\dfrac{t^2+1}{t^2}\cdot\dfrac{\sqrt{1+2t^2}}{|t|}}=-\int\dfrac{|t|\,dt}{(t^2+1)\sqrt{1+2t^2}}$.
   4. For $t>0$: $=-\displaystyle\int\dfrac{t\,dt}{(t^2+1)\sqrt{1+2t^2}}$. Let $w=t^2$, $dw=2t\,dt$: $=-\dfrac12\displaystyle\int\dfrac{dw}{(w+1)\sqrt{2w+1}}$. …

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9. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $f(x)=\displaystyle\int \dfrac{dx}{(x^2+2)}$ and $f(\sqrt2)=0$, then $f(0)=$ (A) $\dfrac{\pi}{2\sqrt2}$ (B) $\dfrac{-\pi}{2\sqrt2}$ (C) $\dfrac{-\pi}{4\sqrt2}$ (D) $\dfrac{\pi}{4\sqrt2}$

   ›Reveal solutionSolution

   Using the standard arctangent antiderivative and the given condition $f(\sqrt2)=0$ to fix the constant of integration, we find $f(0)=-\pi/(4\sqrt2)$.

   Concept and Intuition

   This is an indefinite integral with an added condition that pins down the otherwise-arbitrary constant $C$. We first find the general antiderivative using the standard formula $\int\frac{dx}{x^2+a^2}=\frac1a\arctan\frac{x}{a}+C$, then use the given value at $x=\sqrt2$ to solve for $C$, and finally evaluate at $x=0$.

   Step-by-Step Solution

   1. Recognize the standard form with $a^2=2$, i.e. $a=\sqrt2$:

   $f(x) = \int\frac{dx}{x^2+2} = \frac{1}{\sqrt2}\arctan\left(\frac{x}{\sqrt2}\right)+C$

   2. Apply the condition $f(\sqrt2)=0$:

   $\frac{1}{\sqrt2}\arctan\left(\frac{\sqrt2}{\sqrt2}\right)+C = 0 \implies \frac{1}{\sqrt2}\arctan(1)+C=0$

   Since $\arctan(1)=\pi/4$:

   $\frac{1}{\sqrt2}\cdot\frac{\pi}{4}+C=0 \implies C = -\frac{\pi}{4\sqrt2}$ …

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10. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.If $\displaystyle\int_{0}^{\pi} \dfrac{dx}{1+2\sin^2 x} = k$, then greatest integer less than or equal to k is (A) 2 (B) 0 (C) 1 (D) -1

    ›Reveal solutionSolution

    The integral evaluates to $\pi/\sqrt3\approx1.81$ using the standard $\cos^2+ \text{const}\cdot\sin^2$ formula, so $\lfloor k\rfloor = 1$.

    Concept and Intuition

    Integrals of $\dfrac{1}{a^2\cos^2x+b^2\sin^2x}$ over $[0,\pi]$ have the closed form $\pi/(ab)$ — a standard result obtained by dividing numerator and denominator by $\cos^2x$ and using $\int\sec^2x/(a^2+b^2\tan^2x)dx = \frac1{ab}\tan^{-1}(\frac{b}{a}\tan x)$, doubled appropriately to cover the full range including the discontinuity of $\tan x$ at $\pi/2$.

    Step-by-Step Solution

    1. Rewrite the denominator using $1=\cos^2x+\sin^2x$: $1+2\sin^2x = \cos^2x+\sin^2x+2\sin^2x = \cos^2x+3\sin^2x$.
    2. So $k = \displaystyle\int_0^\pi \frac{dx}{\cos^2x+3\sin^2x} = \int_0^\pi \frac{dx}{1^2\cos^2x + (\sqrt3)^2\sin^2x}$. …

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11. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.$\displaystyle\int_0^{\pi/4} \dfrac{(\cos x - \sin x)}{9 + 5\sin 2x}\, dx =$ (A) $\dfrac{1}{\sqrt{5}}\left(\mathrm{Tan}^{-1}\sqrt{10} - \mathrm{Tan}^{-1}\sqrt{5}\right)$ (B) $\dfrac{1}{2\sqrt{5}}\left(\mathrm{Tan}^{-1}\dfrac{\sqrt{10}}{2} - \mathrm{Tan}^{-1}\dfrac{\sqrt{5}}{2}\right)$ (C) $\dfrac{1}{2\sqrt{5}}\left(\mathrm{Tan}^{-1}\sqrt{10} + \mathrm{Tan}^{-1}\sqrt{5}\right)$ (D) $\dfrac{1}{\sqrt{5}}\left(\mathrm{Tan}^{-1}\sqrt{\dfrac{5}{2}} + \mathrm{Tan}^{-1}\dfrac{\sqrt{5}}{2}\right)$

    ›Reveal solutionSolution

    Substituting $u=\sin x+\cos x$ turns the whole integral into a standard $\int du/(a^2+u^2)$ form; the value is $\dfrac{1}{2\sqrt5}\left(\mathrm{Tan}^{-1}\dfrac{\sqrt{10}}2-\mathrm{Tan}^{-1}\dfrac{\sqrt5}2\right)$.

    Concept and Intuition

    The numerator $\cos x-\sin x$ is (up to sign) exactly the derivative of $\sin x+\cos x$, and $\sin 2x$ can be written in terms of $u=\sin x+\cos x$ via $u^2=1+\sin 2x$. This is the standard trick for integrals mixing $\sin x\pm\cos x$ with $\sin 2x$.

    Step-by-Step Solution

    1. Let $u=\sin x+\cos x$. Then $du=(\cos x-\sin x)\,dx$, and squaring, $u^2=1+2\sin x\cos x=1+\sin 2x$, so $\sin 2x=u^2-1$.
    2. The denominator becomes $9+5\sin2x=9+5(u^2-1)=4+5u^2$.
    3. Limits: at $x=0$, $u=0+1=1$; at $x=\pi/4$, $u=\frac{1}{\sqrt2}+\frac1{\sqrt2}=\sqrt2$.
    4. The integral becomes

    $\int_1^{\sqrt2}\frac{du}{4+5u^2}=\frac15\int_1^{\sqrt2}\frac{du}{u^2+\frac45}.$

    5. Using $\int\dfrac{du}{u^2+k^2}=\dfrac1k\mathrm{Tan}^{-1}\dfrac uk$ with $k=\dfrac{2}{\sqrt5}$: …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{x+1}{(x-2)\sqrt{1-x}}\, dx =$ (A) $\log(x+1) - \log(x-2)\sqrt{1-x} + c$ (B) $\log(x-2)\sqrt{1-x} + c$ (C) $6\tan^{-1}\sqrt{1-x} - 2\sqrt{1-x} + c$ (D) $4\tan^{-1}\sqrt{1-x} - 2\sqrt{1-x} + c$

    ›Reveal solutionSolution

    The substitution $t=\sqrt{1-x}$ turns the rational-irrational integrand into a simple rational function of $t$, giving an arctan plus a linear term in $t$.

    Concept and Intuition

    Whenever an integrand contains only $\sqrt{1-x}$ and rational combinations of $x$, setting $t=\sqrt{1-x}$ (so $x=1-t^2$) clears the square root entirely and converts the whole integral into a rational function of $t$ — a completely mechanical simplification.

    Step-by-Step Solution

    1. Let $t=\sqrt{1-x}$, so $x=1-t^2$, $dx=-2t\,dt$.
    2. Then $x+1=2-t^2$ and $x-2=-1-t^2=-(1+t^2)$.
    3. The integrand becomes:

    $\dfrac{x+1}{(x-2)\sqrt{1-x}}dx=\dfrac{2-t^2}{-(1+t^2)\,t}\cdot(-2t\,dt)=\dfrac{2(2-t^2)}{1+t^2}dt.$

    4. Simplify: $\dfrac{2(2-t^2)}{1+t^2}=\dfrac{4-2t^2}{1+t^2}=\dfrac{-2(1+t^2)+6}{1+t^2}=-2+\dfrac{6}{1+t^2}$. …

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