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 :)

$$
I = \int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx = \int\frac{3x^2-1}{\sqrt{(x + \frac12)^2 - \frac54}}dx
$$
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
$$
I = \frac18 \Big[6 \sqrt{x^2 + x - 1} (2 x - 3) + 13 \tanh^{-1}\frac{2 x + 1}{2 \sqrt{x^2 + x - 1}} \Big] + {\rm constant}
$$
