TITLE: Relation between $\int_0^{\frac{\pi}4}(1+\tan x)^2dx$ and $\int_0^1\frac1{(1+x)^2(1+x^2)}dx$.
QUESTION [8 upvotes]: Let
$$ \displaystyle I_1=\int_0^{\frac{\pi}{4}}(1+\tan x)^2dx, \>\>\>\>\>
\displaystyle I_2=\int_0^{1}\frac{1}{(1+x)^2(1+x^2)}dx$$
then find $\displaystyle\frac{I_1}{I_2}$

My method-
I was able to solve $I_1$ using standard formulae and got $I_1=1+\ln2$. Similarly, I solved for $I_2$ using partial fraction decomposition and got $I_2=0.25 (1+\ln 2)$. Therefore, the required ratio is $4$.
Is there some other way to solve this question?

REPLY [10 votes]: Note
\begin{align}
I_1=\int_0^{\frac{\pi}{4}}(1+\tan x)^2dx
\overset{t=\tan x}= \int_0^{1}\frac{(1+t)^2}{1+t^2}dt
\overset{t\to \frac{1-t}{1+t}}=4I_2
\end{align}