Trig Substitution Integral Question My class is going over trig substitution, but I can't figure this one out, mostly because it's not in the correct form. Could someone help explain how to set up this problem?

$$ \int \frac {1}{x^2-25} dx $$
You don't need trig here: use partial fractions.
The integral
$$ \int \frac {1}{x^2-25} dx $$
isn't in the correct form yet, but you can manipulate it easily
$$ \int \frac {1}{x^2-25} dx =  \int \frac {1}{25((\frac{x}{5})^2-1)} dx = \frac{1}{25}\int \frac {1}{(\frac{x}{5})^2-1} dx$$
now, just substitute $u=\frac{x}{5}, du=\frac{dx}{5}$
$$ \frac{1}{5}\int \frac {1}{u^2-1} du$$
continuing to manipulate...
$$ \frac{1}{5}\int \frac {1}{u^2-1} du = \frac{1}{5}\int \frac {1}{\sec^2(\sec^{-1}u)-1} du=\frac{1}{5}\int \frac {1}{\tan^2(\sec^{-1}u)} du$$
From here on, it's just an inverse trigonometric substitution (albeit a messy one because it involves $\mid u \mid$). It would be far better to use inverse hyperbolic trigonometric substitution in this case, or as Robert Israel has already pointed out, partial fractions.