∫(0到2)xdx/(x∧2－2x+2)∧2答案里有一部步是这样做的，我想问是怎么来化的。

∫(0到2)xdx/(x∧2－2x+2)∧2答案里有一部步是这样做的，我想问是怎么来化的。

x^2-2x+2=(x-1)^2 +1
let
x-1 = tany
dx = (secy)^2 dy
x=0, y =-π/4
x=2, y=π/4

∫(0->2) xdx/(x^2-2x+1)^2
=∫(-π/4->π/4) (secy)^2 (1+tany) dy/(secy)^4
=∫(-π/4->π/4) (1+tany) dy/(secy)^2
=∫(-π/4->π/4) (1+tany)(cosy)^2 dy
=∫(-π/4->π/4)[ (cosy)^2+sinycosy] dy
=(1/2)∫(-π/4->π/4) [1+cos2y+sin2y] dy追问对啊，然后呢？我问的是后面一个等号呀。
仅如图的一步不会。求解～追答∫(0->2) xdx/(x^2-2x+1)^2
=∫(-π/4->π/4) (secy)^2 (1+tany) dy/(secy)^4
=∫(-π/4->π/4) (1+tany) dy/(secy)^2
=∫(-π/4->π/4) (1+tany)(cosy)^2 dy
=∫(-π/4->π/4)[ (cosy)^2+sinycosy] dy
=(1/2)∫(-π/4->π/4) [1+cos2y+sin2y] dy
=(1/4) [2y+sin2y-cos2y] (-π/4->π/4)
=π/4追问你做错了，不过谢谢。
但是我还是不知道我拍的那一步是怎么化的……好吧我自己想吧，谢了。我会做了，谢谢了。