1. ## improper int

how do u do this?
int 0 to a (1/(a^2 - x^2))dx

wut i did was to put c=0
so lim c->0+ int c->a dx/(a^2 - x^2)

am i doing this right?

2. Originally Posted by solars
how do u do this?
int 0 to a (1/(a^2 - x^2))dx

wut i did was to put c=0
so lim c->0+ int c->a dx/(a^2 - x^2)

am i doing this right?
$\displaystyle \int_0^a \frac{1}{x^2 - a^2} \, dx = \lim_{c \rightarrow a} \int_0^c \frac{1}{x^2 - a^2} \, dx$.

3. Originally Posted by mr fantastic
$\displaystyle \int_0^a \frac{1}{x^2 - a^2} \, dx = \lim_{c \rightarrow a} \int_0^c \frac{1}{x^2 - a^2} \, dx$.
just one more question how do u go about integrating 1/(a^2-x^2)?

4. Originally Posted by solars
just one more question how do u go about integrating 1/(a^2-x^2)?
Partial fraction decomposition.

5. Originally Posted by solars
just one more question how do u go about integrating 1/(a^2-x^2)?
Here's an alternative to partial fractions :

$\displaystyle \frac{1}{a^2-x^2}=\frac{1}{(a-x)(a+x)}=\frac{1}{a}\left[\frac{a}{(a-x)(a+x)}\right]=\frac{1}{a}\left[\frac{a+x-x}{(a-x)(a+x)}\right]$ $\displaystyle =\frac{1}{a}\left[\frac{a-x}{(a-x)(a+x)}+\frac{x}{a^2-x^2}\right]=\frac{1}{a}\left[\frac{1}{a+x}-\frac{x}{x^2-a^2}\right]$

So you would have $\displaystyle \frac{1}{a}\left[\int\frac{\,dx}{x+a}-\int\frac{x\,dx}{x^2-a^2}\right]$

--Chris