# Integration Problem

• Mar 23rd 2008, 12:23 PM
Cindi24
Integration Problem
Hello, I'm new here and I have a problem that I can't solve. I thank you in advance for ANY help you can give me. I know things are written to scale, but I couldn't find where to access the symbols. Thank you so much!

∫ x - 1
----------dx
2
x - 4x + 5

this bottom x is squared and the -- lines are the divided by lines. Then the formula is multiplied by dx. Thanks!

Cindi
• Mar 23rd 2008, 12:30 PM
Jhevon
Quote:

Originally Posted by Cindi24
Hello, I'm new here and I have a problem that I can't solve. I thank you in advance for ANY help you can give me. I know things are written to scale, but I couldn't find where to access the symbols. Thank you so much!

∫ x - 1
----------dx
2
x - 4x + 5

this bottom x is squared and the -- lines are the divided by lines. Then the formula is multiplied by dx. Thanks!

Cindi

note that $\displaystyle \int \frac {x - 1}{x^2 - 4x + 5}~dx = \frac 12 \int \frac {2x - 2}{x^2 - 4x + 5}~dx$ $\displaystyle = \frac 12 \int \frac {2x - 4 + 2}{x^2 - 4x + 5}~dx = \frac 12 \int \frac {2x - 4}{x^2 - 4x + 5}~dx + \int \frac 1{x^2 - 4x + 5}~dx$

for the first integral, do a substitution of $\displaystyle u = x^2 - 4x + 5$, for the second, complete the square in the denominator, and try to get the arctangent integral.

can you continue?
• Mar 23rd 2008, 12:31 PM
Jhevon
Quote:

Originally Posted by o_O
$\displaystyle \frac{x-1}{x^{2}-4x+5} = \frac{x-1}{(x+5)(x-1)} = \frac{1}{x+5}$

Easier to deal with?

that's not true. if you expand the denominator of what you wrote, you get $\displaystyle x^2 + 4x - 5$
• Mar 23rd 2008, 12:33 PM
Cindi24
oh, okay! I think so. Thank you!
• Mar 23rd 2008, 12:34 PM
o_O
Ah whoops my bad. Don't know what I was thinking =S
• Mar 23rd 2008, 12:35 PM
Jhevon
Quote:

Originally Posted by o_O
Ah whoops my bad. Don't know what I was thinking =S

hehe, i thought the same thing at first, but then i realized the factorization doesn't work, so i had to do it the hard way