How to proceed $\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $ 
$$I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $$

I have taken the second term of denominator (in square root) as $(x - 2)t$. But cannot go further. Please suggest how to proceed or any alternate method to solve it.
 A: Note that $7x-10-x^2 = \frac94-(\frac72-x)^2$. So, substitute  $\frac72-x=\frac32\sin t$ to integrate as follows
\begin{align}
& \int \frac{dx}{(x-2)(1+\sqrt{7x-10-x^2})}\\
= & - \int \frac{2\cos t}{(1-\sin t)(2+3\cos t)}dt\\
= & - \int \left ( \frac{\cos t}{1-\sin t}-\frac {3\sin t}{2+3\cos t} - \frac {3}{2+3\cos t} \right)dt\\
=& \>\ln(1-\sin t) -\ln(2+3\cos t)+\frac3{\sqrt5}\tanh^{-1}\frac{\tan\frac t2}{\sqrt5}+C
\end{align}
A: We are given $I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)}$
Here we have $7x-10-x^2=(x-2)(5-x)$,
therefore put $\sqrt{7x-10-x^2} =(x-2)t$
on solving and factoring,
we get $x=\frac{\left(2t^2+5\right)}{\left(t^2+1\right)}\ $ therefore
$dx=\frac{-6t}{(t^2+1)^2}$  and $x-2=$ $\frac{(2t^2+5)}{(t^2+1)}-2 =$$\frac{3}{(t^2+1)}$
and $1+\sqrt{7x-10-x^2} = 1+(x-2)t =1+\frac{3t}{t^2+1}$
$=\frac{t^2+3t+1}{t^2+1}$
hence we have $I=\int\frac{t^2+1}{3}.\frac{t^2+1}{t^2+3t+1}.\frac{-6t}{(t^2+1)^2}dt$
$I=-\int\frac{2t}{t^2+3t+1}=-\int\frac{2t+3-3}{t^2+3t+1}dt$
$I=-\int\frac{2t+3}{t^2+3t+1}dt+3\int\frac{dt}{(t+\frac32)^2 -\frac54 }$
$I=ln(t^2+3t+1) -\frac{3}{\sqrt{5}}ln\frac{(2t+3-\sqrt{5})}{(2t+3+\sqrt{5})}$ + C
where $t=\sqrt{\frac{5-x}{x-2}}$.
