How obtain $\sqrt{x^2+x-2} - 3 \ln(\sqrt{x-1} + \sqrt{x+2}) + C$ when integrating $\sqrt{\dfrac{x-1}{x+2}}, x > 1$ The book I'm using states two different answers, one being
$\sqrt{x^2+x-2} - 3 \ln(\sqrt{x-1} + \sqrt{x+2}) + C$ 
I have tried two different approaches.  
Approach one was to rewrite the integral like this:
$\int \sqrt{\dfrac{x-1}{x+2}} dx = \int \dfrac{x-1}{\sqrt{x^2+x-2}} dx = \dfrac{1}{2}\int \dfrac{2x + 1 - 3}{\sqrt{x^2+x-2}} dx = \\ \dfrac{1}{2}\int \dfrac{2x + 1}{\sqrt{x^2+x-2}} dx - \dfrac{3}{2}\int \dfrac{1}{\sqrt{x^2+x-2}} dx = \\
\dfrac{1}{2}[2\sqrt{x^2+x-2} +C_1] - \dfrac{3}{2} [\ln |x + \frac{1}{2} + \sqrt{x^2+x-2}| + C_2] = \\ 
\sqrt{x^2+x-2} - \dfrac{3}{2} \ln |x + \frac{1}{2} + \sqrt{x^2+x-2}|+ C$
Where in the last step I used the fact that $x^2+x-2 = (x+\frac{1}{2})^2 - \frac{9}{4}$ together with the rule that $\int \dfrac{1}{\sqrt{x^2 + a}} dx = \ln|x + \sqrt{x^2+a}|$ 
The result is correct according to the book, but not the form I want to achieve.
Apprach two was to make the substitution $y = \sqrt{\dfrac{x-1}{x+2}} \implies dx = \dfrac{6y}{(y^2-1)^2} dy$
rewriting as
$\int \dfrac{6y^2}{(y^2-1)^2} dy = 6 \int \dfrac{y^2 - 1 + 1}{(y^2-1)^2} dy = 6\int \dfrac{1}{y^2-1} dy + 6 \int \dfrac{1}{(y^2-1)^2} dy$
And using partial fraction decomposition on both integrals. This gave me the answer $\sqrt{x^2+x-2} + \dfrac{3}{2} \ln |x + \frac{1}{2} - \sqrt{x^2+x-2}| + C$ 
Should I try a different approach or am I maybe missing some step in the approaches I've already tried?
 A: In order to prove that
$$\frac{3}{2} \ln (x + \frac{1}{2} - \sqrt{x^2+x-2})=-3 \ln(\sqrt{x-1} + \sqrt{x+2})+C$$
Write
$$-3 \ln{(\sqrt{x-1} + \sqrt{x+2})}$$
$$=3 \ln{(\frac1{\sqrt{x-1} + \sqrt{x+2}})}$$
$$=3 \ln{(\frac{\sqrt{x-1} - \sqrt{x+2}}{(x-1) - (x+2)})}$$
$$=3 \ln{(\frac13(\sqrt{x+2}-\sqrt{x-1}))}$$
$$=3 \ln{(\frac13)}+3\ln{(\sqrt{x+2}-\sqrt{x-1})}$$
$$=-3 \ln{(3)}+\frac32\ln{((\sqrt{x+2}-\sqrt{x-1})^2)}$$
$$=-3 \ln{(3)}+\frac32\ln{((x+2)-2\sqrt{x+2}\sqrt{x-1}+(x-1))}$$
$$=-3 \ln{(3)}+\frac32\ln{(2x+1-2\sqrt{x^2+x-2})}$$
$$=-3 \ln{(3)}+\frac32\ln{(2)}+\ln{(x+\frac12-\sqrt{x^2+x-2})}$$
So as the two functions differ by a constant, they are the same anti derivative - the way in which the result is written does not matter.
A: I would Substitute $$t=\sqrt{\frac{x-1}{x+2}}$$ then $$x=\frac{2t^2+1}{1-t^2}$$ then $$dx=6\,{\frac {t}{ \left( t-1 \right) ^{2} \left( t+1 \right) ^{2}}}dt$$
A: A slightly different substitution from yours gives the result stated in your book:
$$u=\sqrt{x+2}\implies x=u^2-2\implies\mathrm dx=2u\,\mathrm du$$
$$\begin{align*}
\int\sqrt{\frac{x-1}{x+2}}\,\mathrm dx&=\int\frac{\sqrt{u^2-3}}u(2u\,\mathrm du)\\[1ex]
&=2\int\sqrt{u^2-3}\,\mathrm du\\[1ex]
&=u\sqrt{u^2-3}-3\ln\left|\sqrt{u^2-3}+u\right|+C&(*)
\end{align*}$$
Computing $\int\sqrt{u^2-3}\,\mathrm du$ can be indeed be done by parts:
$$\begin{cases}f=\sqrt{u^2-3}\implies\mathrm df=\dfrac u{\sqrt{u^2-3}}\,\mathrm du\\[1ex]\mathrm dg=\mathrm du\implies g=u\end{cases}$$
$$\int\sqrt{u^2-3}\,\mathrm du=u\sqrt{u^2-3}-\int\frac{u^2}{\sqrt{u^2-3}}\,\mathrm du$$
and the remaining integral can be approached with the substitution $u=\sqrt 3\sec t$. (Then again, you could have used this substitution to compute the antiderivative of $\sqrt{u^2-3}$; up to you.)
The result $(*)$ above follows, and replacing $u=\sqrt{x+2}$ gives the final antiderivative,
$$=\underbrace{\sqrt{x+2}\sqrt{x-1}}_{\sqrt{x^2+x-2}}-3\ln\left|\sqrt{x+1}+\sqrt{x+2}\right|+C$$
