分部积分[x^2arctanx⼀(1+x^2)]dx

2025-01-18 20:57:40
推荐回答(1个)
回答1:

∫x^2arctanx/(1+x^2)dx
=∫(x^2+1-1)arctanx/(1+x^2)dx
=∫arctanxdx-∫arctanx/(1+x^2)dx
=xarctanx-∫ x/(1+x^2)dx-∫arctanxd (arctanx)
=xarctanx-(1/2)∫ 1/(1+x^2)d(x^2)-(1/2)(arctanx)^2
=xarctanx-(1/2)ln(1+x^2)-(1/2)(arctanx)^2+C