The value of $\int_{0}^{1} \frac{8 \log (1+x)}{1+x^{2}} d x$ is

JEE Main · 2011 · 10 Apr 2011 · Offline · Definite Integrals

The value of

\int_{0}^{1} \frac{8 l o g ( 1 + x )}{1 + x ^{2}} d x

Answer: (D)

Let

I = \int_{0}^{1} \frac{8 lo g ( 1 + x )}{1 + x ^{2}} d x

and substitute

x = tan θ

, turning it into

8 \int_{0}^{π /4} lo g (1 + tan θ) d θ

. Replacing

θ

\frac{π}{4} - θ

and using

1 + tan (\frac{π}{4} - θ) = \frac{2}{1 + tan θ}

gives

8 \int_{0}^{π /4} lo g \frac{2}{1 + tan θ} d θ = 8 lo g 2 \cdot \frac{π}{4} - I

. Hence

2 I = 2 π lo g 2

and

I = π lo g 2

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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