求不定积分 ∫（ xarctanx）dx=
问题描述：
求不定积分 ∫（ xarctanx）dx=
答
∫（ xarctanx）dx= (1/2)∫arctan(x) d(x^2)
=x^2arctan(x)/2 - (1/2)∫ x^2dx/(1+x^2）
=x^2arctan(x)/2 - (1/2)∫dx + (1/2)∫dx/(1+x^2)
=x^2arctan(x)/2 - (x/2) + arctan(x)/2 + C,
其中，C为任意常数。
答
∫（ xarctanx）dx
=0.5∫（ arctanx）d(x^2+1)
接下来分部积分
答
∫ xarctanx dx
= ∫ arctanx d(x²/2)
= (x²/2)arctanx - (1/2)∫ x² d(arctanx)
= (1/2)x²arctanx - (1/2)∫ x²/(x² + 1) dx
= (1/2)x²arctanx - (1/2)∫ [(x² + 1) - 1]/(x² + 1) dx
= (1/2)x²arctanx - (1/2)∫ dx + (1/2)∫ dx/(x² + 1)
= (1/2)x²arctanx - x/2 + (1/2)arctanx + C