<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] This is an alternative method to compute the given integral.
$$I =\int{\frac{dx}{\sqrt{x^2-3x-10}}} = \int{\frac{dx}{\sqrt{(x-3/2)^2-(7/2)^2}}}\\=\log\left|(x-3/2)+\sqrt{(x-3/2)^2-(7/2)^2}\right|+c.$$
Here $$\int{\frac{dy}{\sqrt{y^2-a^2}}}=\log\left|y+\sqrt{y^2-a^2}\right|+c.$$
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