Question

∫(1,0)xarctan√xdx=

Answer 1

换元，令√x=u，则x=u²，dx=2udu
∫(0---->1) xarctan√xdx
=∫(0---->1) u²arctanu*2udu
=2∫(0---->1) u³arctanudu
=(1/2)∫(0---->1) arctanud(u⁴)
=(1/2)∫(0---->1) arctanud(u⁴)
=(1/2)u⁴arctanu-(1/2)∫(0---->1) u⁴/(1+u²)du
=(1/2)u⁴arctanu-(1/2)∫(0---->1) (u⁴-1+1)/(1+u²)du
=(1/2)u⁴arctanu-(1/2)∫(0---->1) (u²-1)du-(1/2)∫(0---->1) 1/(1+u²)du
=(1/2)u⁴arctanu-(1/2)((1/3)u³-u)-(1/2)arctanu |(0---->1)
=(1/2)(π/4)-(1/2)(-2/3)-(1/2)(π/4)
=1/3

用数学软件验证过的，结果正确。

Answer 2

∫(0→1)xarctan√xdx
=0.5∫(0→1)arctan√xdx²
=0.5x²arctan√x(0→1) - 0.5∫(0→1)x²d(arctan√x)

追问

我求的结果为π/8-1/2+1/2ln2对吗