求定积分x/(x^2-2x+2)^2 上限2 下限0
问题描述：
求定积分x/(x^2-2x+2)^2 上限2 下限0
答
∫xdx/(x^2-2x+2)^2
=∫(x-1)dx/[(x-1)^2+1]^2 +∫d(x-1)/[(x-1)^2+1]^2
(x-1)=tanu
=∫tanudu+∫du
= -ln|cosu|+u+C
=ln|√(x^2-2x+2)|+arctan(x-1)+C
∫[0,2]xdx/[(x^2-2x+2)=ln2+π/4