求下列不定积分,∫xarctanxdx

2024-12-04 09:14:59
推荐回答(2个)
回答1:

∫xarctanxdx
=(1/2)∫arctanxd(x^2)
=(1/2)x^2·arctanx-(1/2)∫x^2d(arctanx)
=(1/2)x^2·arctanx-(1/2)∫[x^2/(1+x^2)]dx
=(1/2)x^2·arctanx-(1/2)∫dx+(1/2)∫[1/(1+x^2)]dx
=(1/2)x^2·arctanx-(1/2)x+(1/2)arctanx+C

回答2:

不适合AA绘声绘色