求定积分∫x^2/(1+x^2)^2dx,上限1,下限0.

问题描述:

求定积分∫x^2/(1+x^2)^2dx,上限1,下限0.

分部积分法也可: x/(1+x^2)^2 dx=-1/2d[1/(1+x^2)]
∫(0→1) x^2/(1+x^2)^2 dx
=-1/2×∫(0→1) x d[1/(1+x^2)]
=-1/4+1/2×∫(0→1) 1/(1+x^2) dx
=-1/4+1/2×arctan1
=-1/4+π/8

设x=tanθ ,0=x^2/(1+x^2)^2=(tanθ)^2*(cosθ)^4=(sinθ)^2(cosθ)^2
dx=dtanθ=dθ/(cosθ)^2
所以原式=∫(sinθ)^2dθ=π/8 -1/4