<s>[INST] 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.
 [/INST] Use: $\int \frac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|+C$
Hence $$\int \frac{dx}{\sqrt{x^2-3x-10}} =  \int \frac{dx}{\sqrt{x^2-3x+\frac{9}{4}-\frac{49}{4}}}=\int \frac{dx}{\sqrt{(x-\frac{3}{2})^2-\frac{49}{4}}} $$
$$=\{x-\frac{3}{2}=t; dx=dt\}= \int \frac{dx}{\sqrt{t^2-\frac{49}{4}}}=\log|t+\sqrt{t^2-\frac{49}{4}}|+C$$
$$=\log|x-\frac{3}{2}+\sqrt{x^2-3x-10}|+C=\log|2x-3+2\sqrt{x^2-3x-10}|+ const$$
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