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 :)

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
$$\frac{3q'}2-\frac{9p'}4+\frac{13r'}8=\frac{3x^2-1}{\sqrt{x^2+x-1}}.$$
