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?
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