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 29 on this concept.

1. CBSE 2025Set 65/4/11 markMCQ
   Q.The value of $\int_0^1 \frac{1}{e^x + e^{-x}}\, dx$ is : (A) $-\frac{\pi}{4}$ (B) $\frac{\pi}{4}$ (C) $\tan^{-1} e - \frac{\pi}{4}$ (D) $\tan^{-1} e$

   ›Reveal solutionSolution

   The integral simplifies by rewriting the denominator as $2\cosh x$, then substituting $t = e^x$ to get a rational function in $t$, which integrates to an arctangent. The value is $\tan^{-1} e - \frac{\pi}{4}$, which is option (C).

   The key insight here is that $e^x + e^{-x}$ is exactly $2\cosh x$, but more usefully, it suggests a substitution that turns the integral into a standard arctangent form. When you see a sum of exponentials in the denominator, your first instinct should be to multiply numerator and denominator by something to simplify — here, multiplying by $e^x$ does the trick.

   Let’s work through it.

   1. Rewrite the integrand Multiply numerator and denominator by $e^x$:

   $\frac{1}{e^x + e^{-x}} = \frac{e^x}{e^{2x} + 1}.$

   This is cleaner because the denominator is now $e^{2x}+1$, which looks like $u^2+1$ after a substitution.

   2. Substitute $t = e^x$ Then $dt = e^x\,dx$, so $dx = \frac{dt}{t}$. But notice: the numerator already has $e^x\,dx$ in disguise. When $x = 0$, $t = e^0 = 1$. When $x = 1$, $t = e^1 = e$. The integral becomes:

   $\int_0^1 \frac{e^x}{e^{2x}+1}\,dx = \int_{1}^{e} \frac{1}{t^2+1}\,dt.$

   That’s a direct substitution — no extra factor needed because $e^x dx = dt$.

   3. Integrate the arctangent form The integral $\int \frac{1}{t^2+1}\,dt$ is $\tan^{-1} t + C$. So:

   $\int_{1}^{e} \frac{1}{t^2+1}\,dt = \left[ \tan^{-1} t \right]_{1}^{e} = \tan^{-1} e - \tan^{-1} 1.$

   4. Evaluate the known arctangent $\tan^{-1} 1 = \frac{\pi}{4}$. Therefore: $$$$

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2. CBSE 2026Set 65/2/11 markMCQ
   Q.If $\int_{0}^{2a} \frac{1}{1+4x^2}\, dx = \frac{\pi}{6}$, then the value of $a$ is (A) $\frac{\sqrt{3}}{4}$ (B) $\frac{\sqrt{3}}{2}$ (C) $\sqrt{3}$ (D) $2\sqrt{3}$

   ›Reveal solutionSolution

   The key idea is to evaluate the definite integral using the standard arctangent formula, set it equal to $\frac{\pi}{6}$, and solve for $a$. The value of $a$ is $\frac{\sqrt{3}}{4}$.

   We are given:

   $\int_{0}^{2a} \frac{1}{1+4x^2}\, dx = \frac{\pi}{6}$

   and need to find $a$ from the options.

   Concept and intuition:

   The integrand $\frac{1}{1+4x^2}$ looks like the derivative of an inverse trigonometric function. Recall that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$. Here, the denominator has $4x^2$ instead of $x^2$, so a substitution or a standard formula adjustment is needed. The standard result is:

   $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C$

   Our denominator is $1 + 4x^2 = 1 + (2x)^2$, so we can treat it as $1^2 + (2x)^2$. This suggests letting $u = 2x$, which will convert the integral into the standard arctangent form.

   Let’s work through it step by step.

   1. Rewrite the integral in a standard form. The denominator is $1 + 4x^2 = 1 + (2x)^2$. So we have:

   $\int \frac{dx}{1 + (2x)^2}$

   This matches $\int \frac{dx}{a^2 + u^2}$ with $a = 1$ and $u = 2x$, but we need to account for the $dx$ vs $du$ change.

   2. Substitute $u = 2x$. Then $du = 2\, dx$, so $dx = \frac{du}{2}$. The limits: when $x = 0$, $u = 0$; when $x = 2a$, $u = 4a$. The integral becomes:

   $\int_{0}^{2a} \frac{1}{1 + (2x)^2}\, dx = \int_{u=0}^{u=4a} \frac{1}{1 + u^2} \cdot \frac{du}{2} = \frac{1}{2} \int_{0}^{4a} \frac{du}{1+u^2}$

   3. Evaluate the standard integral. We know $\int \frac{du}{1+u^2} = \tan^{-1} u + C$. So:

   $\frac{1}{2} \int_{0}^{4a} \frac{du}{1+u^2} = \frac{1}{2} \left[ \tan^{-1} u \right]_{0}^{4a} = \frac{1}{2} \left( \tan^{-1}(4a) - \tan^{-1}(0) \right)$

   Since $\tan^{-1}(0) = 0$, this simplifies to:

   $\frac{1}{2} \tan^{-1}(4a)$

   4. Set equal to the given value and solve. The problem states this equals $\frac{\pi}{6}$:

   $\frac{1}{2} \tan^{-1}(4a) = \frac{\pi}{6}$

   Multiply both sides by 2:

   $\tan^{-1}(4a) = \frac{\pi}{3}$

   Now take the tangent of both sides:

   $$$$

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3. CBSE 2026Set A1 markMCQ
   Q.$\int_0^1\frac{dx}{1 + x^2} =$

   (a) $\frac{\pi}{2}$

   (b) $\frac{\pi}{3}$

   (c) $\frac{\pi}{4}$

   (d) $\pi$

   ›Reveal solutionSolution

   $\int_0^1\tfrac{dx}{1+x^2}=\tan^{-1}1-\tan^{-1}0=\tfrac{\pi}{4}$.

   Since $\int\dfrac{dx}{1+x^2}=\tan^{-1}x$, evaluate between the limits:

   …

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4. CBSE 2026Set A1 markMCQ
   Q.$\int_0^1\frac{x^3\,dx}{1 + x^8} =$

   (a) $\frac{\pi}{2}$

   (b) $\frac{\pi}{4}$

   (c) $\frac{\pi}{8}$

   (d) $\frac{\pi}{16}$

   ›Reveal solutionSolution

   Substitute $t=x^4$: $\int_0^1\tfrac{x^3\,dx}{1+x^8}=\tfrac14\int_0^1\tfrac{dt}{1+t^2}=\tfrac{\pi}{16}$.

   Let $t=x^4$, so $dt=4x^3\,dx$, i.e. $x^3\,dx=\tfrac{dt}{4}$. Limits: $x=0\to t=0$, $x=1\to t=1$. Also $x^8=t^2$. Then

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\int \dfrac{dx}{x^2+16} =$

   (a) $\dfrac{1}{4}\tan^{-1}\left(\dfrac{x}{4}\right)+C$

   (b) $\tan^{-1} x + C$

   (c) $\dfrac{1}{2}\tan^{-1}\left(\dfrac{x}{2}\right)+C$

   (d) None of these

   ›Reveal solutionSolution

   Use the standard integral $\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right)+C$.

   Here $a^2=16 \Rightarrow a=4$.

   …

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6. CBSE 2026Set ANNUAL1 mark
   Q.Evaluate $\displaystyle\int_0^1 \dfrac{dx}{1+x^2}$.

   ›Reveal solutionSolution

   Use the standard integral $\int\frac{dx}{1+x^2}=\tan^{-1}x+C$ and apply the limits.

   $\int_0^1 \frac{dx}{1+x^2} = \Big[\tan^{-1}x\Big]_0^1$

   …

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7. CBSE 2026Set ANNUAL1 mark
   Q.If $\displaystyle\int_{0}^{a}\dfrac{dx}{1+4x^{2}}=\dfrac{\pi}{8}$, then find the value of $a$.

   ›Reveal solutionSolution

   $\int_0^a\frac{dx}{1+4x^2}=\tfrac12\tan^{-1}(2a)=\tfrac{\pi}{8}$ gives $2a=1$, i.e. $a=\tfrac12$.

   Step 1: $\displaystyle\int\frac{dx}{1+(2x)^{2}}=\frac{1}{2}\tan^{-1}(2x)$.

   Step 2: $\displaystyle\int_{0}^{a}\frac{dx}{1+4x^{2}}=\frac{1}{2}\tan^{-1}(2a)-0=\frac{1}{2}\tan^{-1}(2a).$

   …

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8. CBSE 2025Set E1 markMCQ
   Q.$\int \frac{dx}{x^2 + 4} =$

   (a) $\frac{1}{4}\tan^{-1}\frac{x}{4} + k$

   (b) $\frac{1}{2}\tan^{-1}\frac{x}{2} + k$

   (c) $\frac{1}{2}\tan^{-1}\frac{2}{x} + k$

   (d) $2\tan^{-1}\frac{x}{2} + k$

   ›Reveal solutionSolution

   Apply the standard integral $\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac{x}{a}+k$ with $a=2$.

   Here $x^2+4=x^2+2^2$, so $a=2$: …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.∫ (from 0 to 1) dx / (1 + x²) is equal to:

   (a) π/2

   (b) π/4

   (c) π/3

   (d) π/6

   ›Reveal solutionSolution

   $\int \dfrac{dx}{1+x^2} = \tan^{-1}x + C$; evaluate at the limits.

   …

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10. CBSE 2024Set ANNUAL1 mark
    Q.Evaluate $\displaystyle\int \dfrac{1}{x^2+36}\, dx$.

    ›Reveal solutionSolution

    Apply the standard formula $\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac xa+C$ with $a=6$.

    Here $x^2+36 = x^2+6^2$, so $a=6$. Using the standard integral:

    $\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$

    …

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11. CBSE 2024Set D1 markMCQ
    Q.$\int \frac{dx}{a^2 + x^2} =$

    (a) $\frac{1}{a}\tan^{-1}\frac{a}{x} + c$

    (b) $\tan^{-1}\frac{x}{a} + c$

    (c) $\frac{1}{a}\tan^{-1}\frac{x}{a} + c$

    (d) $\frac{1}{a}\tan^{-1} x + c$

    ›Reveal solutionSolution

    Standard form: $\int\dfrac{dx}{a^2+x^2}=\dfrac1a\tan^{-1}\dfrac{x}{a}+c$.

    This is a memorised standard result:

    …

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12. CBSE 2024Set D1 markMCQ
    Q.$\int \frac{x^3\,dx}{1 + x^8} =$

    (a) $\tan^{-1} x^4 + c$

    (b) $4\tan^{-1} x^4 + c$

    (c) $\frac{1}{4}\tan^{-1} x^4 + c$

    (d) $2\tan^{-1} x^4 + c$

    ›Reveal solutionSolution

    Substitute $u = x^4$ so the integral becomes a standard $\tan^{-1}$ form.

    Let $u = x^4 \Rightarrow du = 4x^3\,dx$, so $x^3\,dx = \frac{1}{4}du$. Then …

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