[TEX]I=\int\limits_{-1}^{1}\frac{ln(x^2+1)}{e^x+1}dx[/tex]
[TEX]I=\int\limits_{-1}^{0}\frac{ln(x^2+1)}{e^x+1}dx+\int\limits_{0}^{1}\frac{ln(x^2+1)}{e^x+1}dx=I1+I2(1)[/tex]
xét tích phân
[TEX]I1=\int\limits_{-1}^{0}\frac{ln(x^2+1)}{e^x+1}dx[/tex]
đặt
[TEX]x=-t---->dx=-dt[TEX]
đổi cận
[TEX]x=-1--->t=1;x=0---->t=0[/TEX]
[TEX]I=-\int\limits_{1}^{0}\frac{ln((-t)^2+1)}{e^{-t}+1}dt[/tex]
[TEX]I=\int\limits_{0}^{1}\frac{ln(t^2+1)e^t}{e^t+1}dt=\int\limits_{0}^{1}\frac{ln(x^2+1)e^x}{e^x+1}dx[/tex]
thay vào(1)
[TEX]I=\int\limits_{0}^{1}\frac{ln(x^2+1)(e^x+1)}{e^x+1}dx[/tex]
[TEX]I=\int\limits_{0}^{1}ln(x^2+1)dx[/tex]
đặt
[TEX]u=ln(x^2+1)---->du=\frac{2xdx}{x^2+1}[/TEX]
[TEX]dv=dx---->v=x[/TEX]
[TEX]I=xln(x^2+1)can0--->1-\int\limits_{0}^{1}\frac{2x^2dx}{x^2+1}[/TEX]
[TEX]I=[xln(x^2+1)-2x+arctgx]can0--->1[/TEX]