Question

（x+1）/（x²-2x+5）的不定积分怎么求

Answer 1

x+1= (1/2)(2x-2) +2

∫(x+1)/(x^2-2x+5) dx

=(1/2)∫(2x-1)/(x^2-2x+5) dx + 2∫dx/(x^2-2x+5)

=(1/2)ln|x^2-2x+5| + 2∫dx/(x^2-2x+5)

consider

x^2-2x+5 = (x-1)^2 +4

let

x-1= 2tany

dx= 2(secy)^2 dy

∫dx/(x^2-2x+5)

=(1/2)∫ dy

=(1/2)y + C'

=(1/2)arctan[(x-1)/2] + C'

∫(x+1)/(x^2-2x+5) dx

=(1/2)ln|x^2-2x+5| + 2∫dx/(x^2-2x+5)

=(1/2)ln|x^2-2x+5| + arctan[(x-1)/2] + C

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

8）∫1/(cosx)^2dx=tanx+c

9）∫1/(sinx)^2dx=-cotx+c

10）∫1/√（1-x^2) dx=arcsinx+c

Answer 2

x+1= (1/2)(2x-2) +2

∫(x+1)/(x^2-2x+5) dx
=(1/2)∫(2x-1)/(x^2-2x+5) dx + 2∫dx/(x^2-2x+5)
=(1/2)ln|x^2-2x+5| + 2∫dx/(x^2-2x+5)

consider
x^2-2x+5 = (x-1)^2 +4
let
x-1= 2tany
dx= 2(secy)^2 dy

∫dx/(x^2-2x+5)
=(1/2)∫ dy
=(1/2)y + C'
=(1/2)arctan[(x-1)/2] + C'

∫(x+1)/(x^2-2x+5) dx
=(1/2)ln|x^2-2x+5| + 2∫dx/(x^2-2x+5)
=(1/2)ln|x^2-2x+5| + arctan[(x-1)/2] + C