# Tough Integral (Trig Sub?)

• Apr 17th 2010, 07:04 PM
Audioslavery
Tough Integral (Trig Sub?)
Little help with this one? I can't quite see where its going.

∫2dx/(x^2-2x)

Thanks ya'll
• Apr 17th 2010, 07:12 PM
Gusbob
Quote:

Originally Posted by Audioslavery
Little help with this one? I can't quite see where its going.

∫2dx/(x^2-2x)

Thanks ya'll

Try completing the square for the denominator then use this standard integral:

$\int \frac{du}{u^2-a^2} = \frac{1}{2a} ln \left|\frac{u-a}{u+a} \right| + C$
• Apr 17th 2010, 07:15 PM
dwsmith
Quote:

Originally Posted by Audioslavery
Little help with this one? I can't quite see where its going.

∫2dx/(x^2-2x)

Thanks ya'll

Partial fractions would allow you to integrate without using tables.
• Apr 17th 2010, 08:55 PM
Audioslavery
Quote:

Originally Posted by dwsmith
Partial fractions would allow you to integrate without using tables.

Ah thank you, I used the partial fractions method (although I still lack confidence using that method, algebra was a big fail for me).

I ended up getting -ln|x| + ln|x-2|, I still need to evaluate this from 3 to ∞, I'm a little shaky regarding ∞ and natural log.

so it ends up being [-ln|∞| + ln|∞-2|] - [ln|3| + ln|3-2|]

the final answer is supposedly LN|3| according to my book, am I on the right track?
• Apr 17th 2010, 09:02 PM
dwsmith
Quote:

Originally Posted by Audioslavery
Ah thank you, I used the partial fractions method (although I still lack confidence using that method, algebra was a big fail for me).

I ended up getting -ln|x| + ln|x-2|, I still need to evaluate this from 3 to ∞, I'm a little shaky regarding ∞ and natural log.

so it ends up being [-ln|∞| + ln|∞-2|] - [ln|3| + ln|3-2|]

Before you evaluate, combine your ln functions and then use L'Hopital's Rule.