xarctanxdx在上限1,下限0的 定积分.

∫xarctanxdx
=∫arctanxd(x^2/2)
=(1/2)(x^2)arctanx-(1/2)∫[x^2/(1+x^2)]dx
=(1/2)(x^2)arctanx-(1/2)∫[x^2+1-1]/(1+x^2)dx
=(1/2)(x^2)arctanx-(1/2)∫[1-1/(1+x^2)]dx
=(1/2)(x^2)arctanx-(1/2)x+(1/2)arctanx x从0到1
=(1/2)arctan1-1/2+(1/2)arctan1-(1/2)arctan0
=π/4-1/2

二楼是正确的，shawhom同学忘记xdx=(1/2)d(x^2)中的1/2了

∫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