∫arctanx/x的2次方dx求详解谢谢

Question

∫ arctanx/x的2次方 dx 求详解 谢谢

Answer 1

用分部积分法规Integration by parts:∫vdu = uv- ∫udv
∫ arctanx/x^2 dx
设dv = dx/x^2 ,u= arctanx,则 v = ∫ 1/x^2 dx = -1/x
∫ arctanx/x^2 dx = - arctanx/x - ∫ (-1/x) d(arctanx)
= - arctanx/x+ ∫ (1/x)(1/(1+x^2) dx
= -arctanx/x+ ∫ (1/[x(1+x^2) ] dx
= -arctanx/x+ ∫ (x^2-x^2+1)/[x(1+x^2) ] dx
= -arctanx/x+ ∫ (x^2+1)/[x(1+x^2) ] - (x^2)/[x(1+x^2) ] dx
= -arctanx/x+ ∫ [ 1/x - (x)/(1+x^2) ] dx
= -arctanx/x+ ∫ 1/x dx - ∫ (x)/(1+x^2) ] dx
= -arctanx/x + lnx - (1/2) ln (1+x^2) +c