# Thread: need help with intergration

1. ## need help with intergration

I'm stuck on intergrating a function. the function is (x-2)/(x^2-4x+5)^.5. The problem states that we are to use (x-2)^2 as u for subsition. So i took the derivative of (x-2)^2 and plugged du into the intergal. Then I'm stuck. I have .5*the intergal of du/(u+1)^.5. I don't know what to do after that. Any help?

2. Originally Posted by flash9286
I'm stuck on intergrating a function. the function is (x-2)/(x^2-4x+5)^.5. The problem states that we are to use (x-2)^2 as u for subsition. So i took the derivative of (x-2)^2 and plugged du into the intergal. Then I'm stuck. I have .5*the intergal of du/(u+1)^.5. I don't know what to do after that. Any help?
You took care of hardest part!

.5*INT 1/sqrt(u + 1) du

let u + 1 = v --> dv = du

.5*INT 1/sqrt(v) dv
.5*INT (v)^(-1/2)
.5*2*(v)^(1/2) + C
= sqrt(v) + C
= sqrt(u + 1) + C
= sqrt[(x - 2)^2 + 1] + C

3. Thanks for your help.

4. Hello, flash!

They surely suggested that substitution to give you practice,
. . but there is a simpler way . . .

. . . . . . . . . . .(x - 2) dx
We have: . ∫ --------------- . = .
(x² - 4x + 5)^{-½} (x - 2) dx
. . . . . . . . . √x² - 4x + 5

Let: .u .= .x² - 4x + 5 . . . . du .= .(2x - 4)dx . . . . dx .= .du/2(x-2)

Substitute and we have: .½
u^{-½} du