Question

求x/（x²-2x+2）²不定积分

Answer 1

(1/2)arctan(x-1)+(1/2)x(x-1)/(x^2-2x+2)+C

解题过程如下：

I = ∫xdx/(x^2-2x+2)^2 = ∫xdx/[(x-1)^2+1]^2,

令 x-1= tant， 则 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

记作∫f(x)dx或者∫f（高等微积分中常省去dx），即∫f(x)dx=F(x)+C。其中∫叫做积分号，f(x)叫做被积函数，x叫做积分变量，f(x)dx叫做被积式，C叫做积分常数或积分常量，求已知函数的不定积分的过程叫做对这个函数进行不定积分。

扩展资料

常用积分公式：

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

Answer 2

I = ∫xdx/(x^2-2x+2)^2 = ∫xdx/[(x-1)^2+1]^2,
令 x-1= tant， 则 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