∫ln(x^2+1)dx,怎么算
问题描述：
∫ln(x^2+1)dx,怎么算
答
分部积分法：
=xln(x^2+1)-∫xdln(x^2+1)
=xln(x^2+1)-2∫x^2/(x^2+1)dx
=xln(x^2+1)-2∫(1-1/(x^2+1))
=xln(x^2+1)-(2x-arctanx)+C
答
∫ln(x^2+1)dx=∫xln(x^2+1)/xdx=∫ln(x^2+1)/xdx^2=∫xd1/(x^2+1)=x/(x^2+1)+∫1/(x^2+1)dx=arctanx
答
分部积分∫ln(x^2+1)dx = ∫x d ln(x^2+1) = xln(x^2+1) - ∫x d ln(x^2+1)= xln(x^2+1) - 2∫(x^2/x^2+1)dx= xln(x^2+1) - 2∫(x^2+1-1)/(x^2+1)dx= xln(x^2+1) - 2[∫(x^2+1)/(x^2+1)dx -∫(1/x^2+1)dx]= xln(x^2+...