∫xarctanxdx

Answer 1

∫xarctanxdx
=(1/2)∫ arctanxd(x²)
那么使用分部积分法得到，
=(1/2)x²arctanx - (1/2)∫ x²/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ (x²+1-1)/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ 1 dx + (1/2)∫ 1/(1+x²) dx
=(1/2)x²arctanx - (1/2)x + (1/2)arctanx + C，C为常数

Answer 2

令a=1即可，详情如图所示