Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$

Compute

$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$

Answer

Put $x = \tan\theta$, then your integral transforms to $$I= \int_{0}^{\pi/4} \log(1+\tan\theta) \ d\theta. \tag{1}$$

Now using the property that $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx,$$ we have $$I = \int_{0}^{\pi/4} \log\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} \log\biggl(\frac{2}{1+\tan\theta} \biggr) \ d\theta.\tag{2}$$

Adding $(1)$ and $(2)$ we get $$2I = \int_{0}^{\pi/4} \log(2) \ d\theta\Rightarrow I= \log(2) \cdot \frac{\pi}{8}.$$

Attribution
Source : Link , Question Author : user 1591719 , Answer Author : amWhy

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