Evaluate $\displaystyle \int \limits_{0}^{1} \dfrac{\log \left(1 + t\right)}{1 + t^2} \; dt$
Let $I = \displaystyle \int \limits_{0}^{1} \dfrac{\log \left(1 + t\right)}{1 + t^2} \; dt$ $\;\;\; \cdots \; (1)$
Put $\tan^{-1}\left(t\right) = x$ $\;\;\; \cdots \; (2)$
$\implies$ $t = \tan \left(x\right)$ $\;\;\; \cdots \; (2a)$
Differentiating equation $(2)$ gives
$\dfrac{dt}{1 + t^2} = dx$ $\;\;\; \cdots \; (2b)$
$\text{When } \begin{cases}
t = 0, & x = \tan^{-1} \left(0\right) = 0 \\
t = 1, & x = \tan^{-1} \left(1\right) = \dfrac{\pi}{4}
\end{cases}$ $\;\;\; \cdots \; (2c)$
$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ can be written as
$\begin{aligned}
I & = \int \limits_{0}^{\frac{\pi}{4}} \log \left[1 + \tan x\right] \; dx \\\\
& = \int \limits_{0}^{\frac{\pi}{4}} \log \left[1 + \tan \left(\dfrac{\pi}{4} - x\right)\right] \; dx \;\;\; \cdots \; (3) \hspace{1em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(a - x\right) \; dx\right]
\end{aligned}$
$\begin{aligned}
\text{Now, } \tan\left(\dfrac{\pi}{4} - x\right) & = \dfrac{\tan \left(\dfrac{\pi}{4}\right) - \tan \left(x\right)}{1 + \tan \left(\dfrac{\pi}{4}\right) \times \tan \left(x\right)} \\\\
& = \dfrac{1 - \tan \left(x\right)}{1 + \tan \left(x\right)} \\\\
\therefore \; 1 + \tan \left(\dfrac{\pi}{4} - x\right) & = 1 + \dfrac{1 - \tan \left(x\right)}{1 + \tan \left(x\right)} \\\\
& = \dfrac{1 + \tan \left(x\right) + 1 - \tan \left(x\right)}{1 + \tan \left(x\right)} \\\\
& = \dfrac{2}{1 + \tan x} \;\;\; \cdots \; (4)
\end{aligned}$
$\therefore$ $\;$ In view of equation $(4)$ equation $(3)$ becomes
$I = \displaystyle \int \limits_{0}^{\frac{\pi}{4}} \log \left[\dfrac{2}{1 + \tan x}\right] \; dx$
i.e. $I = \displaystyle \int \limits_{0}^{\frac{\pi}{4}} \log \left(2\right) \; dx - \int \limits_{0}^{\frac{\pi}{4}} \log \left(1 + \tan x\right) \; dx$
i.e. $I = \log \left(2\right) \displaystyle \int \limits_{0}^{\frac{\pi}{4}} dx - I$
i.e. $2 I = \log \left(2\right) \times \left[x\right]_{0}^{\pi / 4}$
i.e. $2 I = \dfrac{\pi}{4} \log \left(2\right)$
i.e. $I = \dfrac{\pi}{8} \log \left(2\right)$