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

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Definite Integrals

The value of the integral

\int_{- 1}^{1} (\frac{x ^{3} + ∣ x ∣ + 1}{x ^{2} + 2∣ x ∣ + 1}) d x

is equal to

Answer: (B)

The denominator is

x^{2} + 2∣ x ∣ + 1 = (∣ x ∣ + 1)^{2}

. The term

\frac{x ^{3}}{( ∣ x ∣ + 1 ) ^{2}}

is odd, so it integrates to

0

over

[- 1, 1]

. The term

\frac{∣ x ∣ + 1}{( ∣ x ∣ + 1 ) ^{2}} = \frac{1}{∣ x ∣ + 1}

is even, so

I = 2 \int_{0}^{1} \frac{d x}{x + 1} = 2 [ln (x + 1)]_{0}^{1} = 2 ln 2

. Hence the integral equals

2 lo g_{e} 2

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