# Calc 2 : Integrals

• Feb 18th 2009, 11:09 AM
Coco87
Calc 2 : Integrals
Hey,
It's most likely my poor Algebra, but I'm having an issue with this problem:

Solve the integral: $\int{\frac{x-1}{x^2-4x+5}}dx$

$\int{\frac{x-1}{x^2-4x+5}}dx \Longrightarrow \int{\frac{x-1}{(x-2)^2+(3)^2}}dx$

$\Longrightarrow \int{\frac{(x-2)+1}{(x-2)^2+9}}dx$

$\Longrightarrow \int{\frac{u+1}{u^2+9}}dx$

where: $u=x-2$ and $du=dx$

$\Longrightarrow \int{\frac{u}{u^2+9}}du + \int{\frac{1}{u^2+9}}du$

set: $p=u^2+9$ and $\frac{1}{2}dp=udu$

$\Longrightarrow \frac{1}{2}\int{\frac{1}{p}}dp + \int{\frac{1}{u^2+9}}du$

$= \frac{1}{2}\ln{(x^2-4x+5)}+\frac{1}{3}\arctan{\left( \frac{x-2}{3}\right)}+C$

The problem is, the book shows:

$= \frac{1}{2}\ln{(x^2-4x+5)}+\arctan{(x-2)}+C$

So where did the $\frac{1}{3}$('s) go?

I've done the problem 3 times and can't figure out what I've done wrong. Any ideas?
• Feb 18th 2009, 11:17 AM
running-gag
Hi

$(x-2)^2+(3)^2 = x^2 - 4x + 13$
• Feb 18th 2009, 12:56 PM
Coco87
Quote:

Originally Posted by running-gag
Hi

$(x-2)^2+(3)^2 = x^2 - 4x + 13$