HD:
Đặt [TEX]x=tant[/TEX]\Rightarrow[TEX]dx=(1+tan^2t)dt[/TEX]
[tex]I=\int\limits_{0}^{1}\frac{ln(1+x)}{1+x^2}dx=[/tex][TEX]\int\limits_{0}^{\frac{\pi}{4}}{ln(1+tant)}dt[/TEX]
Đặt[TEX] u=\frac{\pi}{4}-t[/TEX]\Rightarrow[TEX]du=-dt[/TEX]
[tex]I=\int\limits_{0}^{\frac{\pi}{4}}{ln(1+tan({\frac{\pi}{4}-u))}}du[/tex]
[tex]=\int\limits_{0}^{\frac{\pi}{4}}{ln(\frac{2}{1+tanu})du[/tex]
[tex]=\int\limits_{0}^{\frac{\pi}{4}}ln2du-I[/tex]
Bạn tính tiếp nhé!