Question

∫ln(x^2+1)dx，怎么算

∫ln(x^2+1)dx，怎么算

Answer 1

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+1) - 2[∫dx -∫(1/x^2+1)dx]
= xln(x^2+1) - 2[x - arctanx]+C

Answer 2

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

Answer 3

∫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