∫xarctanxdx=1/2 ∫arctanxdx^2
=1/2[x^2arctanx|(0,1)-∫(0,1)x^2/(1+x^2)dx]
=1/2[π/4-∫(0,1)1-1/(1+x^2)dx]
=1/2[π/4-∫(0,1)dx+∫(0,1)1/(1+x^2)dx]
=1/2[π/4-x|(0,1)+arctanx|(0,1)]
=π/4-1/2
∫xarctanxdx
分部积分！
=x^2arctanx-∫x^2/(1+x^2)dx
=x^2arctanx-∫1-1/(1+x^2)dx
=x^2arctanx-∫1dx+∫1/(1+x^2)dx
=x^2arctanx-x+arctanx
因为上限1，下限0
所以。代入得
π/2-1
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024