神马作文网∫（0到2）xdx/(x^2-2x+2)^2 令x=1+tanu 后得出 原式=2 ∫(0到π /4)（cosu）^2d
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∫（0到2）xdx/(x^2-2x+2)^2 令x=1+tanu 后得出 原式=2 ∫(0到π /4)（cosu）^2du?
∫[0,2]xdx/(x^2-2x+2)^2
=∫[0,2] (1/2)(2x-2)dx/(x^2-2x+2)^2+∫[0,2]dx/(x^2-2x+2)^2
=(1/2)∫[0,2]d(x^2-2x+2)/(x^2-2x+2)^2 +∫[0,2]d(x-1)/[(x-1)^2+1]^2
=(-1/2)(1/(x^2-2x+2))|[0,2] +(1/2)arctan(x-1)|[0,2] +(1/2)(x-1)/(x^2-2x+2)|[0,2]
=(1/2)*(π/2）+1/2
..x=1+tanu,d(x-1)=secu^2du
∫d(x-1)/[(x-1)^2+1]^2=∫cosu^2du=(1/2)∫(1+cos2u)du=(1/2)u+(1/2)sinucosu
=(1/2)arctan(x-1)+(1/2)(x-1)/(x^2-2x+2)
=∫[0,2] (1/2)(2x-2)dx/(x^2-2x+2)^2+∫[0,2]dx/(x^2-2x+2)^2
=(1/2)∫[0,2]d(x^2-2x+2)/(x^2-2x+2)^2 +∫[0,2]d(x-1)/[(x-1)^2+1]^2
=(-1/2)(1/(x^2-2x+2))|[0,2] +(1/2)arctan(x-1)|[0,2] +(1/2)(x-1)/(x^2-2x+2)|[0,2]
=(1/2)*(π/2）+1/2
..x=1+tanu,d(x-1)=secu^2du
∫d(x-1)/[(x-1)^2+1]^2=∫cosu^2du=(1/2)∫(1+cos2u)du=(1/2)u+(1/2)sinucosu
=(1/2)arctan(x-1)+(1/2)(x-1)/(x^2-2x+2)
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