Question

Calculate the antiderivative of dx/(x^2-9)

Answer 1

We have
\[\int \frac{dx}{x^2-9}\]\
this can be written as
\[ \int \frac{dx}{(x+3)(x-3)}\]
Let's multiply and divide the numerator by 6
\[ \int \frac{6dx}{6(x+3)(x-3)}\]
Now 6 can be written as
\[ 6=(x+3)-(x-3)\]
Now the numerator can be replaced by (x+3)-(x-3)
we get
\[ \int \frac{(x+3)-(x-3)}{6(x+3)(x-3)}dx\]
We get now
\[ \int \frac{dx}{6(x-3)}- \frac{dx}{6(x+3)}\]
Now performing integration
\[ \frac{1}{6}\ln(x-3)-\frac{1}{6}\ln(x+3)+c\]
Log terms can be combined , we get
\[\frac{1}{6}\ln{\frac{x-3}{x+3}}+c\]

Answer 2

...Very clever, Ash. I was going to suggest a trigonometric substitution of sin(x), but looks like you gots it, yo.

Answer 3

Thanks Badreferences:D

Answer 4

I'm GOING to have to remember that trick.

Answer 5

Caro did you understand it?

Answer 6

If the asker is curious, if you use the identity cos(x)^2=1-sin(x)^2, and you use x=nsin(x), you can solve this a la trigonometric substitution.

Answer 7

Yup that was a perfect/clear explanation. Thans a lot!!