Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$ Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$
I solved this integral by euler substitution by replacing
$\sqrt{x^2+x-1}=x+t$
but it's not allowed by the problem.
p.s Is there any other method to solve with?
Thank you in advance :)
 A: $$
I = \int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx = \int\frac{3x^2-1}{\sqrt{(x + \frac12)^2 - \frac54}}dx
$$
Replacing  $x= y\cdot \sqrt{\frac54} - \frac12$ gives
$$
I =  \int\frac{(15 y^2)/4 - (3 \sqrt 5  y)/2 - 1/4}{\sqrt{(y^2-1)}}dy
$$
Now let $y = \cosh (z)$ which gives
$$
I =  \int [15 (\cosh (z))^2)/4 - (3 \sqrt 5  \cosh (z))/2 - 1/4 ] \; dz
$$
This can easily be performed, giving
$$I =  \frac{13 z - 12 \sqrt 5 \sinh(z) + 15 \sinh(z) \cosh(z)}{8} + {\rm constant}$$
Replacing $z$ by $x$ (via $y$) gives
$$
I = \frac18 \Big[6 \sqrt{x^2 + x - 1} (2 x - 3) + 13 \tanh^{-1}\frac{2 x + 1}{2 \sqrt{x^2 + x - 1}} \Big] + {\rm constant}
$$
A: Hint:
We observe that
$$p':=\left(\sqrt{x^2+x-1}\right)'=\frac{2x+1}{2\sqrt{x^2+x-1}},$$
$$q':=\left(x{\sqrt{x^2+x-1}}\right)'=\frac{4x^2+3x-2}{2\sqrt{x^2+x-1}}.$$
Also, by completing the square,
$$x^2+x-1=\left(x+\frac12\right)^2-\frac54=\frac54\left(\left(\frac{2x+1}{\sqrt5}\right)^2-1\right)$$
and
$$r':=\left(\text{arcosh}\frac{2x+1}{\sqrt5}\right)'=\frac1{\sqrt{x^2+x-1}}$$
Hence we form a linear combination to match the numerator of the integrand and we find
$$\frac{3q'}2-\frac{9p'}4+\frac{13r'}8=\frac{3x^2-1}{\sqrt{x^2+x-1}}.$$
A: Note
\begin{align}
\int \sqrt{x^2+x-1} \>dx
&\overset{IBP}=\frac12 x \sqrt{x^2+x-1} +\frac14\int \frac {x-2}{\sqrt{x^2+x-1}}dx\\
\end{align}
Then
\begin{align}
&\int\frac{3x^2-1}{\sqrt{x^2+x-1}}\>dx\\
=& 3 \int \sqrt{x^2+x-1} dx - \int \frac {3x-2}{\sqrt{x^2+x-1}}dx\\
=& \frac32 x \sqrt{x^2+x-1} -\frac98 \int \frac {2x+1}{\sqrt{x^2+x-1}}dx + \frac{13}8 \int \frac {1}{\sqrt{x^2+x-1}}dx \\
 =& \frac32 x \sqrt{x^2+x-1} -\frac94 \sqrt{x^2+x-1}
+ \frac{13}8 \cosh^{-1}\frac{2x+1}{\sqrt5}+C
\end{align}
