JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Definite Integral of $\frac{\ln (x)}{x^{2} + a^{2}}$ on the Semi-Infinite Interval

The value of the integral integral from 0 to ∞ of (log_e(x))/(x^2 + 4) dx is:

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A(π log_e(2))/2B(π log_e(2))/4C(1 + π log_e(2))/2D(2 + π log_e(2))/2

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A · (π log_e(2))/2

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

B · (π 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 · (1 + π 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 · (2 + π 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 $\int_{0}^{\infty} \frac{1}{x^{2} + 4} d x$ without constant offsets.

Reviewed route

Solution

StepWorking

01Given

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

02Approach

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

04Execute

For $I_{1} = \int_{0}^{\infty} \frac{\ln t}{t^{2} + 1} d t$ , substitute $t = \frac{1}{u}, d t = - \frac{d u}{u^{2}}$ : $I_{1} = \int_{\infty}^{0} \frac{- \ln u}{\frac{1}{u^{2}} + 1} (- \frac{d u}{u^{2}}) = - \int_{0}^{\infty} \frac{\ln u}{u^{2} + 1} d u = - I_{1} ⟹ 2 I_{1} = 0 ⟹ I_{1} = 0$

05Execute

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

✓Verify

For general form $\int_{0}^{\infty} \frac{\ln x}{x^{2} + a^{2}} d x = \frac{π \ln a}{2 a}$ . For $a = 2$ , this yields $\frac{π \ln 2}{2 (2)} = \frac{π \ln 2}{4}$ .

Hints that build this answer step by step

Let $I = \int_{0}^{\infty} \frac{\ln (x)}{x^{2} + 4} d x$ . 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?
$I = \int_{0}^{\infty} \frac{\ln (4) - \ln (t)}{t^{2} + 4} d t$
Add $I = \int_{0}^{\infty} \frac{\ln (x)}{x^{2} + 4} d x$ and $I = \int_{0}^{\infty} \frac{\ln (4) - \ln (t)}{t^{2} + 4} d t$ . What is the resulting value of $2 I$ ?
$2 I = \ln (4) \int_{0}^{\infty} \frac{1}{t^{2} + 4} d t$
Evaluate the integral $\int_{0}^{\infty} \frac{1}{t^{2} + 4} d t$ and solve for $I$ . What is $I$ ?
$I = \frac{π \ln (2)}{4}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $\frac{\ln (2)}{2} \int_{0}^{\infty} \frac{1}{t^{2} + 4} d t$ , which evaluates to $\frac{π \ln (2)}{4}$ .

Why each option works or fails

A: (π log_e(2))/2 - The student correctly simplifies to $2 I = \ln (2) \int_{0}^{\infty} \frac{d t}{t^{2} + 4}$ , but forgets to divide by 2 when solving for $I$ , or computes $\int_{0}^{\infty} \frac{d t}{t^{2} + 4}$ as $π$ instead of $\frac{π}{4}$ . Remember that $\int_{0}^{\infty} \frac{d t}{t^{2} + a^{2}} = {[\frac{1}{a} \arctan (\frac{t}{a})]}_{0}^{\infty} = \frac{π}{2 a}$ , and ensure both sides are divided by the coefficient of $I$ .
B · correct: (π 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: (1 + π 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: (2 + π 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 $\int_{0}^{\infty} \frac{1}{x^{2} + 4} d x$ without constant offsets.

Step-by-step solution

Given: Integral $I = \int_{0}^{\infty} \frac{\ln x}{x^{2} + 4} d x$
Approach: Substitute $x = 2 t$ 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 $x = 2 t ⟹ d x = 2 d t$ . When $x = 0, t = 0$ and as $x \to \infty, t \to \infty$ : $I = \int_{0}^{\infty} \frac{\ln (2 t)}{4 t^{2} + 4} (2 d t) = \frac{1}{2} \int_{0}^{\infty} \frac{\ln 2 + \ln t}{t^{2} + 1} d t = \frac{\ln 2}{2} \int_{0}^{\infty} \frac{1}{t^{2} + 1} d t + \frac{1}{2} \int_{0}^{\infty} \frac{\ln t}{t^{2} + 1} d t$
Execute: For $I_{1} = \int_{0}^{\infty} \frac{\ln t}{t^{2} + 1} d t$ , substitute $t = \frac{1}{u}, d t = - \frac{d u}{u^{2}}$ : $I_{1} = \int_{\infty}^{0} \frac{- \ln u}{\frac{1}{u^{2}} + 1} (- \frac{d u}{u^{2}}) = - \int_{0}^{\infty} \frac{\ln u}{u^{2} + 1} d u = - I_{1} ⟹ 2 I_{1} = 0 ⟹ I_{1} = 0$
Execute: Compute the remaining constant term integral: $I = \frac{\ln 2}{2} {[\tan^{- 1} t]}_{0}^{\infty} = \frac{\ln 2}{2} \cdot \frac{π}{2} = \frac{π \ln 2}{4}$
Verify: For general form $\int_{0}^{\infty} \frac{\ln x}{x^{2} + a^{2}} d x = \frac{π \ln a}{2 a}$ . For $a = 2$ , this yields $\frac{π \ln 2}{2 (2)} = \frac{π \ln 2}{4}$ .

Definite Integral of $\frac{\ln (x)}{x^{2} + a^{2}}$ on the Semi-Infinite Interval

What feels right?

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

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

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