JEE Main 2026 Apr 6 Shift 2, Mathematics Q20: Some Properties of Definite Integrals

JEE Main2026Apr 6, Shift 2Mathematics

The value of the integral $\int_{- 1}^{1} (\frac{x ^{3} + ∣ x ∣ + 1}{x ^{2} + 2∣ x ∣ + 1}) d x$ is equal to:

A
$3 lo g_{e} 2$
B
$2 lo g_{e} 2$
C
$5 lo g_{e} 3$
D
$3 lo g_{e} 3$

Solution

Split the integrand. The denominator $x^{2} + 2∣ x ∣ + 1 = (∣ x ∣ + 1)^{2}$ is even. The term $\frac{x ^{3}}{( ∣ x ∣ + 1 ) ^{2}}$ is odd, so its integral over $[- 1, 1]$ vanishes.

Remaining: $\int_{- 1}^{1} \frac{∣ x ∣ + 1}{( ∣ x ∣ + 1 ) ^{2}} d x = \int_{- 1}^{1} \frac{1}{∣ x ∣ + 1} d x$ .

By symmetry: $= 2 \int_{0}^{1} \frac{1}{x + 1} d x = 2 [ln (x + 1)]_{0}^{1} = 2 ln 2$ .

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JEE Main 2026 Apr 6 Shift 2, Mathematics Q20: Some Properties of Definite Integrals

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