\(\int_0^af(x)dx=\int_0^af(a-x)dx \\ \text{Let I}=\int_0^{\pi/4} log(1+tanx)dx=\int_0^{\pi/4}log(1+tan[{\pi/4}-x]dx \\= \int_0^{\pi/4}log(1+(1-tanx)/(1+tanx))dx = \int_0^{\pi/4}log(2/(1+tanx))dx \\= \int_0^{\pi/4}log2dx-\int_0^{\pi/4}log(1+tanx)dx = \int_0^{\pi/4}log2dx-I \\2I=\int_0^{\pi/4}log2dx \ and \ I=log2*{\pi/4}*1/2\\={\pi/8}\ log2\)
