The integral $ \int_{0}^{\pi/4} \frac {\ln (1+\tan x)}{(1+\tan x)^2} dx? $

Question

In order to produce a solution of $I$, in

Integrating $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx$ with restricted tehniques

for a teacher's delight, I landed up in the integral $$J=\int_{0}^{\pi/4} \frac {\ln (1+\tan x)}{(1+\tan x)^2} dx? $$ Here is how $J$ emeges. $$I=\int_{0}^{1}\frac{x\ln(1+x)}{1+x^2}dx=\int_{0}^{\pi/4} \tan t\ln(1+\tan t) dt$$ $$=\int_{0}^{\pi/4} \tan (\pi/4-t)\ln(1+\tan(\pi/4- t)) ~dt= \int_{0}^{\pi/4} \frac{1-\tan t}{1+\tan t} [\ln 2 -\ln(1+\tan t)] dt =$$ $$\int_{0}^{\pi/4} \frac{2-\mbox{sec}^2 t}{(1+\tan t)^2}[\ln 2-\ln(1+\tan t)]=I_1+I_2+I_3+I_4.$$ Here $$I_1=2 \ln 2\int_{0}^{\pi/4} \frac{dt}{(1+\tan t)^2}=\ln 2\int_{0}^{\pi/4} \frac{1+\cos 2t}{(1+\sin 2 t)}dt= \ln 2\int_{0}^{\pi/4} \frac{1-\sin 2t}{(1-\sin^ 2 2t)}dt+$$ $$\frac{1}{2} \ln 2 ~\ln((1+\sin 2t)|_{0}^{\pi/2}= \lim_{t\rightarrow \pi/4}( \tan 2t- \mbox{sec} 2t)\frac{\ln 2}{2}+\frac{ \ln^2 2}{2} =\frac{1}{2}(\ln 2+\ln^2 2).$$ $$I_2=- \ln 2 \int_{0}^{\pi/4} \frac{\mbox{sec}^2 t ~ dt}{(1+\tan t)^2}=- \frac{1}{2}\ln 2.$$ $$I_3=\int_{0}^{\pi/4} \frac{\mbox{sec}^2 t \ln(1+\tan t) ~ dt}{(1+\tan t)^2}=\int_{0}^{\ln 2} u e^{-u} ~du=\frac{1}{2}(1-\ln 2). $$ $$I_4=-2\int_{0}^{\pi/4} \frac {\ln (1+\tan t)}{(1+\tan t)^2} dt =-2 J.$$ So, here the question is : How to solve the $J$ integral? The expected answer is: $$J=\frac{3\ln^2 2}{16}-\frac{\pi^2}{192}+\frac{1-\ln 2}{4}.$$ This comes by using the value of $I$.But, here the idea is not to use it.

Answer 1

With $u=\tan(x)$ we get $$J=\int_{0}^{\pi/4} \frac {\ln (1+\tan x)}{(1+\tan x)^2} dx = \int_0^1 \frac{du}{u^2+1} \frac{\ln(1+u)}{(1+u)^2}$$ Then Integration by parts, where we differentiate the logarithm $$J=\ln(1+u)\frac{1}{4}\left(2\ln(1+u)-\frac{2}{u+1}-\ln(1+u^2)\right)|_0^1 +\\ \frac{1}{4} \int_0^1du \frac{1}{1+u}\left(\ln(1+u^2) +\frac{2}{u+1}-2\ln(1+u)\right) \\= \frac{\ln(2)}{4}(2\ln(2)-1-\ln(2))+\frac{1}{4}(I_1+I_2-I_3)$$

Here, $I_2$ and $I_3$ are very easy. Use $x=1+u$ for $I_2$ to obtain $$I_2=1$$ And use integration by parts for $I_3$, where you differentiate the logarithm to get $$I_3=2\ln(2)^2-I_3 \Rightarrow I_3=\ln(2)^2$$ The integral $I_1$ is tricky, but the solution is given here to be $$I_1=-\frac{\pi^{2}}{48} + \frac{3}{4}\ln(2)^2$$

Combining everything we obtain $$J=\frac{3\ln^2 2}{16}-\frac{\pi^2}{192}+\frac{1-\ln 2}{4}$$ as stated by you.

Edit: I noticed that the $I_1$ integral is too close to the integral $I$ that you forbid to use... Sorry about that.