Find the in tegration1

Calculus Level 4

0 1 x ln ( x + 1 ) x 2 + 1 d x \large \int_0^1 \dfrac{x \ln(x + 1)}{x^2+1} \, dx

Find the closed form of the integral above and submit your answer to 3 decimal places.


The answer is 0.162.

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1 solution

Chew-Seong Cheong
Jan 26, 2018

I = 0 1 x ln ( x + 1 ) x 2 + 1 d x Let x = tan θ d x = sec 2 θ = 0 π 4 tan θ ln ( tan θ + 1 ) d θ = 0 π 4 tan θ ln ( cos θ + sin θ cos θ ) d θ = 0 π 4 sin θ ln ( 2 cos ( θ π 4 ) ) cos θ d θ 0 π 4 sin θ ln ( cos θ ) cos θ d θ \begin{aligned} I & = \int_0^1 \frac {x\ln(x+1)}{x^2+1}dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \\ & = \int_0^\frac \pi 4 \tan \theta \ln (\tan \theta +1)\ d \theta \\ & = \int_0^\frac \pi 4 \tan \theta \ln \left(\frac {\cos \theta + \sin \theta}{\cos \theta} \right) d \theta \\ & = \int_0^\frac \pi 4 \frac {\sin \theta \ln \left(\sqrt 2\cos \left(\theta - \frac \pi 4\right)\right)}{\cos \theta} \ d \theta - \int_0^\frac \pi 4 \frac {\sin \theta \ln \left(\cos \theta\right)}{\cos \theta} \ d \theta \end{aligned}

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