Calc 2 : Integrals

**Coco87** · February 18th 2009, 11:09 AM

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?

**running-gag** · February 18th 2009, 11:17 AM

Hi

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

**Coco87** · February 18th 2009, 12:56 PM

Originally Posted by running-gag

Hi

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

Arrrrrgggg!!!

I knew I did something stupid...

Well, thank you for your correction

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