1樓：匿名使用者
(1/2)arctan(x-1)+(1/2)x(x-1)/(x^2-2x+2)+c
解題過程如下：
i = ∫xdx/(x^2-2x+2)^2 = ∫xdx/[(x-1)^2+1]^2,
令 x-1= tant， 則 x=1+tant， dx=(sect)^2dt,
i = ∫xdx/[(x-1)^2+1]^2 = ∫(1+tant)dt/(sect)^2
= ∫[(cost)^2+sintcost]dt
= ∫[1/2+(1/2)cos2t+sintcost]dt
= t/2+(1/4)sin2t+(1/2)(sint)^2+c
= (1/2)arctan(x-1)+(1/2)x(x-1)/(x^2-2x+2)+c
記作∫f(x)dx或者∫f（高等微積分中常省去dx），即∫f(x)dx=f(x)+c。其中∫叫做積分號，f(x)叫做被積函式，x叫做積分變數，f(x)dx叫做被積式，c叫做積分常數或積分常量，求已知函式的不定積分的過程叫做對這個函式進行不定積分。
常用積分公式：
1）∫0dx=c
2）∫x^udx=(x^(u+1))/(u+1)+c
3）∫1/xdx=ln|x|+c
4）∫a^xdx=(a^x)/lna+c
5）∫e^xdx=e^x+c
6）∫sinxdx=-cosx+c
7）∫cosxdx=sinx+c
2樓：匿名使用者
i = ∫xdx/(x^2-2x+2)^2 = ∫xdx/[(x-1)^2+1]^2,
令 x-1= tant， 則 x=1+tant， dx=(sect)^2dt,
i = ∫xdx/[(x-1)^2+1]^2 = ∫(1+tant)dt/(sect)^2
= ∫[(cost)^2+sintcost]dt
= ∫[1/2+(1/2)cos2t+sintcost]dt
= t/2+(1/4)sin2t+(1/2)(sint)^2+c
= (1/2)arctan(x-1)+(1/2)x(x-1)/(x^2-2x+2)+c
求不定積分∫x/（x²+2x+2）dx
3樓：匿名使用者
∫x/（
x²+2x+2）dx=1/2*∫(2x+2-2)/（x²+2x+2）dx
=1/2*∫(2x+2)/（x²+2x+2）dx-∫1/（x²+2x+1+1）dx
=1/2ln（x²+2x+2）+arctan(x+1)+c
求不定積分∫x/(x^2+2x+2)dx
4樓：匿名使用者
解∫x/(x²+2x+2)dx
=1/2∫(2x+2-2)/(x²+2x+2)dx=1/2∫(2x+2)/(x²+2x+2)dx-∫1/(x²+2x+2)dx
=1/2∫1/(x²+2x+2)d(x²+2x+2)-∫1/[(x+1)²+1]dx
=1/2∫1/udu-∫1/[(x+1)²+1]d(x+1)=1/2ln|u|-∫1/(u²+1)du=1/2ln(x²+2x+2)-acrtanu+c=1/2ln(x²+2x+2)-arctan(x+1)+c
5樓：匿名使用者
答：∫[x/(x^2+2x+2)]dx
=∫ dx
=∫d(x+1) - ∫ d(x+1)
=(1/2)∫ d[(x+1)²+1] - ∫ d(x+1)=(1/2) ln [(x+1)²+1] -arctan(x+1)+c
= ln√(x²+2x+2) -arctan(x+1)+c
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