Question

计算下列积分，求解答过程

Answer 1

详细过程在这里，希望能够帮到你，望采纳哦…………

追答

第二题修改一下

在拍个清楚的

Answer 2

(1)

x-x^2 = 1/4-(x-1/2)^2
let
x-1/2 = (1/2)sinu
dx = (1/2)cosu du
x=0, u=-π/2
x=1, u=π/2
∫(0->1) dx/√(x-x^2)
=∫(-π/2->π/2) du
=π
(2)

∫(0->1) ln(1+x)/(2-x)^2 dx
=∫(0->1) ln(1+x) d[1/(2-x)]
=[ ln(1+x)/(2-x) ]|(0->1) - ∫(0->1) dx/[(2-x)(1+x)]
=ln2 + ∫(0->1) dx/[(x-2)(1+x)]
=ln2 + (1/3)∫(0->1) [1/(x-2)-1/(1+x)] dx
=ln2 +(1/3) [ln|(x-2)/(x+1)|]|(0->1)
=ln2 +(1/3) [ ln( 1/2) - ln2 ]
=ln2 -(2/3)ln2
=(1/3)ln2