# Thread: Calc 2 : Integrals

1. ## Calc 2 : Integrals

Hey,
It's most likely my poor Algebra, but I'm having an issue with this problem:

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

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

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

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

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

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

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

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

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

The problem is, the book shows:

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

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

I've done the problem 3 times and can't figure out what I've done wrong. Any ideas?

2. Hi

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

3. Originally Posted by running-gag
Hi

$\displaystyle (x-2)^2+(3)^2 = x^2 - 4x + 13$
Arrrrrgggg!!!

I knew I did something stupid...

Well, thank you for your correction