The value of is :

Engineering

Mathematics

Properties of Definite Integral

Question

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

πlog2

$\frac{π}{8} \log 2$

log2

$\frac{π}{2} \log 2$

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x = tanq

dx = sec²q dq

$I = \int_{0}^{π / 4} \frac{3 \ln (1 + tanθ)}{\sec^{2} θ} \sec^{2} dθ = 8 \int_{0}^{π / 4} \ln (1 + tanθ) dθ$

$\begin{matrix} \Rightarrow I = 8 \int_{0}^{π / 4} \ln (1 + \tan (π / 4 - θ)) dθ \\ = 8 \int_{0}^{π / 4} \ln (\frac{2}{1 + tanθ}) dθ \\ \Rightarrow 2 I = 8 \int_{0}^{π / 4} \ln (2) dθ \\ = 8 \frac{π}{4} \ln 2 = 2 πln 2 \end{matrix}$

⇒ l = π ln2