Question

xarctanxdx在上限1，下限0的 定积分。要过程

Answer 1

∫腔迟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/好圆耐2[π/4-∫(0,1)dx+∫(0,1)1/(1+x^2)dx]
=1/友春2[π/4-x|(0,1)+arctanx|(0,1)]
=π/4-1/2

Answer 2

∫xarctanxdx
分部积分！腔简慧
=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
所以咐亩。代伍答入得
π/2-1

Answer 3

∫xarctanxdx
=∫肢喊arctanxd(x^2/纯唤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/历裤野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

Answer 4

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