Evaluate $\int_{0}^{\infty} \frac{\ln (1+u) -\ln 2}{(u+1)\sqrt{u} \ln u} du$ Please help me with evaluate the following improper integral $$\int_{0}^{\infty} \frac{\ln (1+u) -\ln 2}{(u+1)\sqrt{u} \ln u} du.$$
 A: Let the considered integral be $I$. Use the substitution $u=\tan^2\theta \Rightarrow du=2\tan\theta \sec^2\theta\,d\theta$ to obtain:
$$I=\int_0^{\pi/2} \frac{\ln(1+\tan^2\theta)-\ln 2}{(1+\tan^2\theta)\tan \theta \ln(\tan^2\theta)}2\tan\theta\sec^2\theta\,d\theta$$
Since $1+\tan^2\theta=\sec^2\theta$,
$$I=\int_0^{\pi/2} \frac{2\ln(\sec\theta)-\ln 2}{\sec^2\theta \tan\theta (2\ln(\tan\theta))}2\tan\theta\sec^2\theta\,d\theta=\int_0^{\pi/2} \frac{-2\ln(\cos\theta)-\ln 2}{\ln(\tan\theta)}d\,\theta\,\,\,\,\,(*)$$
Using the property of definite integrals which states that $\int_a^bf(x)dx=\int_a^b f(a+b-x)\,dx$, we can write:
$$I=\int_0^{\pi/2} \frac{-2\ln(\sin\theta)-\ln 2}{\ln(\cot\theta)}\,d\theta=\int_0^{\pi/2}\frac{2\ln(\sin\theta)+\ln 2}{\ln(\tan\theta)}\,d\theta\,\,\,\,\,\,\,(**)$$
Add $(*)$ and $(**)$ to obtain:
$$2I=\int_0^{\pi/2} \frac{2\ln(\sin\theta)-2\ln(\cos\theta)}{\ln(\tan\theta)}\,d\theta=\int_0^{\pi/2} \frac{2\ln(\tan\theta)}{\ln(\tan\theta)}\,d\theta$$
$$\Rightarrow I=\int_0^{\pi/2}\,d\theta=\boxed{\dfrac{\pi}{2}}$$
