Question

Sol. [Key Concept: Substitute $x = 2t$; the $ln\ t$ part cancels by symmetry, leaving $I = \frac{\pi ln\ 2}{4}$.]

Put $x = 2t \Rightarrow dx = 2dt$

$${I = \int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}n2t}{4t^{2} + 4}(2dt) = \frac{1}{2}\int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}n2 + \mathcal{l}nt}{t^{2} + 1}dt }{= \frac{1}{2}\int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}n2}{t^{2} + 1}dt + \frac{1}{2}\int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}nt}{t^{2} + 1}dt }{\left\lbrack \frac{ln2}{2}\tan^{- 1}t \right\rbrack_{0}^{\infty} + I_{1} }{= \frac{\mathcal{l}n2}{2} \cdot \frac{\pi}{2} + I_{1} }{I_{2} = \frac{1}{2}\int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}nt}{t^{2} + 1}dt\ t = \frac{1}{u} \Rightarrow dt = - \frac{du}{u^{2}} }{= \frac{1}{2}\int_{\infty}^{0}\mspace{2mu}\frac{\mathcal{l}n(1/u)}{\frac{1}{u^{2}} + 1}\left( - \frac{du}{u^{2}} \right) }$$Add $\Rightarrow I_{1} = 0$

$$I = \frac{\pi\mathcal{l}n2}{4} $$