First slide

Evaluation of definite integrals

Question

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

Moderate

A

log 2
B

$π \log 2$
C

$\frac{π}{8} \log 2$
D

$\frac{π}{2} \log 2$

Solution

Let $I = 8 \int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} d x$
$= 8 \int_{0}^{π / 4} \log \frac{(1 + \tan θ)}{\sec^{2} θ} \sec^{2} θ d θ (x = \tan θ)$
So $I = 8 \int_{0}^{π} \log (1 + \tan θ) d θ$
$\begin{array}{l} = 8 \int_{0}^{π / 4} \log (1 + \tan (\frac{π}{4} - θ) d θ \\ (\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x) \\ = 8 \int_{0}^{π / 4} \log (1 + \frac{1 - \tan θ}{1 + \tan θ}) d θ \\ = 8 \int_{0}^{π / 4} \log \frac{2}{1 + \tan θ} d θ = (8 \log 2) \frac{π}{4} - I \end{array}$
$\Rightarrow 2 I = (8 \log 2) \frac{π}{4} \Rightarrow I = π \log 2$

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