I'm back! With a nice integral for you

There are a couple ways to do this (manipulating with sum to product formulas was one way a friend used) but this is my preferred method.

$I=\displaystyle\int_0^\frac{\pi}{4}\ln(1+\tan x)\text{ }dx$

We can use the special case of the identity $\displaystyle\int_a^b f(x)\text{ }dx=\int_a^b f(a+b-x)\text{ }dx$ saying that $\displaystyle\int_0^kf(x)\text{ }dx=\displaystyle\int_0^kf(k-x)\text{ }dx.$

$\begin{array}{c}f\int_0^\frac{\pi}{4}\ln(1+\tan x)\text{ }dx&=\int_0^\frac{\pi}{4}\ln\left(1+\tan\left(\dfrac{\pi}{4}-x\right)\right)\text{ }dx\\ &=\int_0^\frac{\pi}{4}\ln\left(1+\dfrac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\right)\text{ }dx\\ &=\int_0^\frac{\pi}{4}\ln\left(1+\dfrac{1-\tan x}{1+\tan x}\right)\text{ }dx\\ &=\int_0^\frac{\pi}{4}\ln\left(\dfrac{1+\tan x}{1+\tan x}+\dfrac{1-\tan x}{1+\tan x}\right)\text{ }dx\\ &=\int_0^\frac{\pi}{4}\ln\left(\dfrac{2}{1+\tan x}\right)\text{ }dx\\ &=\int_0^\frac{\pi}{4}\ln2-\ln(1+\tan x)\text{ }dx\\ &=\int_0^\frac{\pi}{4}\ln2\text{ }dx-\int_0^\frac{\pi}{4}\ln(1+\tan x)\text{ }dx\\ &=\dfrac{\pi\ln2}{4}-I \end{array}$

Bringing the $I$ to the left side, we find that $2I=\dfrac{\pi\ln2}{4}\Rightarrow I=\boxed{\dfrac{\pi\ln2}{8}}$

I = ∫(0,π/4)ln(1 + tanx)dx

I = ∫(0,π/4)ln(1+tan(π/4-x)) dx

= ln(1+(tanπ/4-tanx)/(1+tanπ/4tanx))

= ln (1+(1-tanx)/(1+tanx))

= ln(2/(1+tanx))

On adding, 2I = ∫(0,π/4)ln2 dx

2I =ln2 ∫(0,π/4)dx

I = ((ln2)/2) ［x］(0,π/4)

I = (πln2)/8

$I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \ln { (1+tanx)dx\quad \quad (Applying\quad \int _{ a }^{ b }{ f(x)dx=\int _{ a }^{ b }{ f(a+b-x)dx) } } } } \\ I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(1+tan(\frac { \pi }{ 4 } -x } ))dx\\ =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(1+\frac { 1-tanx }{ 1+tanx } } )dx\\ =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(\frac { 1+tanx }{ 1+tanx } +\frac { 1-tanx }{ 1+tanx } } )dx$

$I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(\frac { 2 }{ 1+tanx } } )dx\\ =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln2\quad -\quad ln(1+tanx)\quad dx } \\ =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln2\quad dx\quad -\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(1+tanx)dx } } \\ =ln2\left[ \frac { \pi }{ 4 } \right] \quad -\quad I\\ 2I=\frac { \pi }{ 4 } ln2\\ I=\frac { \pi }{ 8 } ln2$