Interesting Integrals 5

Substitute $x= \tan u \Rightarrow dx = \sec^2 u \, du$ and the integral becomes $I=8 \int _{ 0 }^{ { \pi }/{ 4 } }{ \cfrac { \ln { \left( 1+\tan { u } \right) } }{ 1+\tan ^{ 2 }{ u } } } \cdot \sec ^{ 2 }{ u } \, du = 8 \int _{ 0 }^{ { \pi }/{ 4 } }{ \ln { \left( 1+\tan { u } \right) } } \, du .$ This is a familiar integral for me as I have posted this problem in the past.

Using $\displaystyle \int _{ a }^{ b }{ f(u) } \, du=\int _{ a }^{ b }{ f(a+b-u) } \, du$ , we can note that $2I=8 \int _{ 0 }^{ { \pi }/{ 4 } }{ \ln { \left\{ \left( 1+\tan { u } \right) \left( 1+\tan { \left( \cfrac { \pi }{ 4 } -u \right) } \right) \right\} } } \, du = 8 \int _{ 0 }^{ { \pi }/{ 4 } }{ \ln { 2 } } \, du = 2 \pi \ln 2 .$

Therefore $I= \pi \ln 2 \Rightarrow A=2 \Rightarrow A^2 = \large \color{#20A900}{\boxed{4}}$ .