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Integral Calculus: JEE Main Mathematics Question with Solution

The value of the integral

\int_{0}^{\infty} \frac{\log_e(x)}{x^2 + 4} dx

is :

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What feels right?

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\frac{\pi \log_e(2)}{2}

\frac{\pi \log_e(2)}{4}

\frac{1 + \pi \log_e(2)}{2}

\frac{2 + \pi \log_e(2)}{2}

Step-by-step solutionView

Correct answer

Using the substitution $x = \frac{4}{t}$ , the integral simplifies to $\frac{\ln(2)}{2} \int_{0}^{\infty} \frac{1}{t^2+4} dt$ , which evaluates to $\frac{\pi \ln(2)}{4}$ .

Option analysis

Why each option works or fails

A ·

\frac{\pi \log_e(2)}{2}

The student correctly simplifies to $2I = \ln(2) \int_{0}^{\infty} \frac{dt}{t^2 + 4}$ , but forgets to divide by 2 when solving for $I$ , or computes $\int_{0}^{\infty} \frac{dt}{t^2 + 4}$ as $\pi$ instead of $\frac{\pi}{4}$ . Remember that $\int_{0}^{\infty} \frac{dt}{t^2 + a^2} = \left[ \frac{1}{a} \arctan\left(\frac{t}{a}\right) \right]_0^\infty = \frac{\pi}{2a}$ , and ensure both sides are divided by the coefficient of $I$ .

B ·

\frac{\pi \log_e(2)}{4}

Correct solution. This is the correct value obtained by substituting $x = 4/t$ and using symmetry to eliminate the self-cancelling logarithmic term.

C ·

\frac{1 + \pi \log_e(2)}{2}

The student improperly integrates $\ln(x)$ by parts without taking advantage of inversion symmetry, leading to an extraneous non-zero boundary term like $[\frac{1}{2}\arctan(x/2)\ln(x)]$ evaluated incorrectly. Avoid integration by parts directly on unbounded logarithmic intervals; instead, substitute $x = a^2/t$ to exploit reciprocal symmetry.

D ·

\frac{2 + \pi \log_e(2)}{2}

The student introduces an extraneous constant term from misapplying the substitution limits or adding boundary values incorrectly during algebraic manipulation. Check that the transformation $x = 4/t$ maps $(0, \infty)$ onto $(\infty, 0)$ , cleanly producing a multiple of the standard integral $\int_0^\infty \frac{1}{x^2+4}dx$ without constant offsets.

Reviewed route

Solution

StepWorking

01given

Integral $I = \int_{0}^{\infty} \frac{\ln x}{x^2 + 4} dx$

02approach

Substitute $x = 2t$ to rescale the denominator to the standard form $t^2 + 1$ , splitting the logarithm into $\ln 2 + \ln t$ . The integral of $\frac{\ln t}{t^2 + 1}$ vanishes by reciprocal substitution $t = 1/u$ .

03execute

Substitute $x = 2t \implies dx = 2dt$ . When $x=0, t=0$ and as $x\to\infty, t\to\infty$ : $I = \int_{0}^{\infty} \frac{\ln(2t)}{4t^2 + 4} (2dt) = \frac{1}{2} \int_{0}^{\infty} \frac{\ln 2 + \ln t}{t^2 + 1} dt = \frac{\ln 2}{2} \int_{0}^{\infty} \frac{1}{t^2 + 1} dt + \frac{1}{2} \int_{0}^{\infty} \frac{\ln t}{t^2 + 1} dt$

04execute

For $I_1 = \int_{0}^{\infty} \frac{\ln t}{t^2 + 1} dt$ , substitute $t = \frac{1}{u}, dt = -\frac{du}{u^2}$ : $I_1 = \int_{\infty}^{0} \frac{-\ln u}{\frac{1}{u^2} + 1} \left(-\frac{du}{u^2}\right) = -\int_{0}^{\infty} \frac{\ln u}{u^2 + 1} du = -I_1 \implies 2I_1 = 0 \implies I_1 = 0$

05execute

Compute the remaining constant term integral: $I = \frac{\ln 2}{2} \left[ \tan^{-1} t \right]_{0}^{\infty} = \frac{\ln 2}{2} \cdot \frac{\pi}{2} = \frac{\pi \ln 2}{4}$

✓verify

For general form $\int_0^\infty \frac{\ln x}{x^2+a^2} dx = \frac{\pi \ln a}{2a}$ . For $a = 2$ , this yields $\frac{\pi \ln 2}{2(2)} = \frac{\pi \ln 2}{4}$ .

Hints that build this answer step by step

Let $I = \int_{0}^{\infty} \frac{\ln(x)}{x^2 + 4} dx$ . Which substitution exploits the reciprocal symmetry of the integrand?
$x = \frac{4}{t}$
Under the substitution $x = \frac{4}{t}$ , what does the integral $I$ become?
$I = \int_{0}^{\infty} \frac{\ln(4) - \ln(t)}{t^2 + 4} dt$
Add $I = \int_0^\infty \frac{\ln(x)}{x^2+4}dx$ and $I = \int_0^\infty \frac{\ln(4)-\ln(t)}{t^2+4}dt$ . What is the resulting value of $2I$ ?
$2I = \ln(4) \int_{0}^{\infty} \frac{1}{t^2 + 4} dt$
Evaluate the integral $\int_{0}^{\infty} \frac{1}{t^2 + 4} dt$ and solve for $I$ . What is $I$ ?
$I = \frac{\pi \ln(2)}{4}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why does the integral of ln(t)/(1+t^2) vanish on (0, inf)?

Under the mapping t -> 1/t, ln(t) becomes -ln(t) while dt/(1+t^2) transforms to -dt/(1+t^2). The reversal of integration limits cancels the minus sign on the differential, leaving the net integrand multiplied by -1 over the same interval.

Answer

Using the substitution $x = \frac{4}{t}$ , the integral simplifies to $\frac{\ln(2)}{2} \int_{0}^{\infty} \frac{1}{t^2+4} dt$ , which evaluates to $\frac{\pi \ln(2)}{4}$ .

Why each option works or fails

A: $\frac{\pi \log_e(2)}{2}$ - The student correctly simplifies to $2I = \ln(2) \int_{0}^{\infty} \frac{dt}{t^2 + 4}$ , but forgets to divide by 2 when solving for $I$ , or computes $\int_{0}^{\infty} \frac{dt}{t^2 + 4}$ as $\pi$ instead of $\frac{\pi}{4}$ . Remember that $\int_{0}^{\infty} \frac{dt}{t^2 + a^2} = \left[ \frac{1}{a} \arctan\left(\frac{t}{a}\right) \right]_0^\infty = \frac{\pi}{2a}$ , and ensure both sides are divided by the coefficient of $I$ .
B · correct: $\frac{\pi \log_e(2)}{4}$ - Correct solution. This is the correct value obtained by substituting $x = 4/t$ and using symmetry to eliminate the self-cancelling logarithmic term.
C: $\frac{1 + \pi \log_e(2)}{2}$ - The student improperly integrates $\ln(x)$ by parts without taking advantage of inversion symmetry, leading to an extraneous non-zero boundary term like $[\frac{1}{2}\arctan(x/2)\ln(x)]$ evaluated incorrectly. Avoid integration by parts directly on unbounded logarithmic intervals; instead, substitute $x = a^2/t$ to exploit reciprocal symmetry.
D: $\frac{2 + \pi \log_e(2)}{2}$ - The student introduces an extraneous constant term from misapplying the substitution limits or adding boundary values incorrectly during algebraic manipulation. Check that the transformation $x = 4/t$ maps $(0, \infty)$ onto $(\infty, 0)$ , cleanly producing a multiple of the standard integral $\int_0^\infty \frac{1}{x^2+4}dx$ without constant offsets.

Step-by-step solution

given: Integral $I = \int_{0}^{\infty} \frac{\ln x}{x^2 + 4} dx$
approach: Substitute $x = 2t$ to rescale the denominator to the standard form $t^2 + 1$ , splitting the logarithm into $\ln 2 + \ln t$ . The integral of $\frac{\ln t}{t^2 + 1}$ vanishes by reciprocal substitution $t = 1/u$ .
execute: Substitute $x = 2t \implies dx = 2dt$ . When $x=0, t=0$ and as $x\to\infty, t\to\infty$ : $I = \int_{0}^{\infty} \frac{\ln(2t)}{4t^2 + 4} (2dt) = \frac{1}{2} \int_{0}^{\infty} \frac{\ln 2 + \ln t}{t^2 + 1} dt = \frac{\ln 2}{2} \int_{0}^{\infty} \frac{1}{t^2 + 1} dt + \frac{1}{2} \int_{0}^{\infty} \frac{\ln t}{t^2 + 1} dt$
execute: For $I_1 = \int_{0}^{\infty} \frac{\ln t}{t^2 + 1} dt$ , substitute $t = \frac{1}{u}, dt = -\frac{du}{u^2}$ : $I_1 = \int_{\infty}^{0} \frac{-\ln u}{\frac{1}{u^2} + 1} \left(-\frac{du}{u^2}\right) = -\int_{0}^{\infty} \frac{\ln u}{u^2 + 1} du = -I_1 \implies 2I_1 = 0 \implies I_1 = 0$
execute: Compute the remaining constant term integral: $I = \frac{\ln 2}{2} \left[ \tan^{-1} t \right]_{0}^{\infty} = \frac{\ln 2}{2} \cdot \frac{\pi}{2} = \frac{\pi \ln 2}{4}$
verify: For general form $\int_0^\infty \frac{\ln x}{x^2+a^2} dx = \frac{\pi \ln a}{2a}$ . For $a = 2$ , this yields $\frac{\pi \ln 2}{2(2)} = \frac{\pi \ln 2}{4}$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Using the substitution $x = \frac{4}{t}$ , the integral simplifies to $\frac{\ln(2)}{2} \int_{0}^{\infty} \frac{1}{t^2+4} dt$ , which evaluates to $\frac{\pi \ln(2)}{4}$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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Students also ask

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