Integrating a Rational Function with a Square Root : $ \int \frac{dx}{\sqrt{x^2-3x-10}} $ 
Integrate $$ \int \frac{dx}{\sqrt{x^2-3x-10}}. $$

I started off with the substitution $\sqrt{x^2-3x-10} = (x-5)t$. To which I got
$$ x = \frac{2+5t^2}{t^2-1} \implies \sqrt{x^2-3x-10} = \left(\frac{2+5t^2}{t^2-1} - 5\right)t = \frac{7t}{t^2-1} $$
and 
$$ dx = \frac{-14t}{(t^2-1)^2} dt. $$
Substituting back in
\begin{align*}
 \int \frac{t^2-1}{7t} \frac{-14t}{(t^2-1)^2}dt &= -2 \int \frac{dt}{t^2-1} \\
&= - \ln\left| \frac{t-1}{t+1} \right| +c \\ 
&= - \ln\left| \frac{\frac{\sqrt{x^2-3x-10}}{x-5} - 1}{\frac{\sqrt{x^2-3x-10}}{x-5} + 1} \right| +c \\
&= - \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| + c.
\end{align*}
Is this a valid answer? Wolfram Alpha's answer is slightly different. I know this can come from a difference in method and/or the functions differ by a constant, but with an expression like this it's hard to check.

See
$$- \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| = \ln\left| \frac{\sqrt{x^2-3x-10} + x - 5}{\sqrt{x^2-3x-10} - x + 5} \right| $$
and
\begin{align*}
\frac{\sqrt{x^2-3x-10} + x - 5}{\sqrt{x^2-3x-10} - x + 5}
=&
\frac{\sqrt{x^2-3x-10} + (x - 5)}{\sqrt{x^2-3x-10} - (x - 5)}
\times
\frac{\sqrt{x^2-3x-10} + (x - 5)}{\sqrt{x^2-3x-10} + (x - 5)}
\\
=&
\frac{2x^2-13x+15+2(x - 5)\sqrt{x^2-3x-10}  }{7(x - 5)}
\\
=&
\frac{2x-3+2\sqrt{x^2-3x-10}  }{7}
\end{align*}
so
$$- \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| = \ln|2x-3+2\sqrt{x^2-3x-10} |-\ln7$$
