The value of the integral $\int\limits_{0}^{\frac{π}{4}}\log_e (1+\tan x)dx$ is:

$\frac{π}{16}\log_e4$

$I=\int\limits_{0}^{\frac{π}{4}}\log (1+\tan x)dx$

$⇒I=\int\limits_{0}^{\frac{π}{4}}\log(1+\tan(\frac{π}{4}-x))dx$

$[∵\int\limits_{0}^af(x)dx=\int\limits_{0}^af(a−x)dx]$

$\tan(\frac{π}{4}-x)=\frac{\tan\frac{π}{4}-\tan x}{1+\tan\frac{π}{4}\tan x}$

$=\frac{1-\tan x}{1+\tan x}$

$I=\int\limits_{0}^{\frac{π}{4}}\log(1+\frac{(1-\tan x)}{(1+\tan x)})dx$

$I=\int\limits_{0}^{\frac{π}{4}}\log(\frac{1+\tan x+1-\tan x}{1+\tan x})dx$

$I=\int\limits_{0}^{\frac{π}{4}}\log(\frac{2}{1+\tan x})dx⇒I=\int\limits_{0}^{\frac{π}{4}}\log 2-\int\limits_{0}^{\frac{π}{4}}\log(1+\tan x)dx$

$I=\log 2\int\limits_{0}^{\frac{π}{4}}dx-I$

$⇒2I=\frac{π}{4}\log 2⇒I=\frac{π}{8}\log 2=\frac{π}{16}\log 4$