# Thread: Tough Integral (Trig Sub?)

1. ## Tough Integral (Trig Sub?)

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

∫2dx/(x^2-2x)

Thanks ya'll

2. 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:

$\displaystyle \int \frac{du}{u^2-a^2} = \frac{1}{2a} ln \left|\frac{u-a}{u+a} \right| + C$

3. 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.

4. 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?

5. 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.