Question

用分部积分法求 ∫arctanxdx 写明白u和v是谁

Answer 1

u=arctanx,v=x

u'=1/(1+x^2),v'=1

原式=∫arctanx×(x)'dx

=∫uv'dx

=uv-∫u'vdx

=xarctanx-∫[x/(1+x^2)]dx

（下面使用凑微分法）

=xarctanx-(1/2)∫[1/(1+x^2)]d(1+x^2)

=xarctanx-(1/2)ln(1+x^2)+C