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.
 A: 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 $$\int{\frac{dy}{\sqrt{y^2-a^2}}}=\log\left|y+\sqrt{y^2-a^2}\right|+c.$$
A: 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
$$- \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| = \ln|2x-3+2\sqrt{x^2-3x-10} |-\ln7$$
A: Use: $\int \frac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|+C$
Hence $$\int \frac{dx}{\sqrt{x^2-3x-10}} =  \int \frac{dx}{\sqrt{x^2-3x+\frac{9}{4}-\frac{49}{4}}}=\int \frac{dx}{\sqrt{(x-\frac{3}{2})^2-\frac{49}{4}}} $$
$$=\{x-\frac{3}{2}=t; dx=dt\}= \int \frac{dx}{\sqrt{t^2-\frac{49}{4}}}=\log|t+\sqrt{t^2-\frac{49}{4}}|+C$$
$$=\log|x-\frac{3}{2}+\sqrt{x^2-3x-10}|+C=\log|2x-3+2\sqrt{x^2-3x-10}|+ const$$
