How can one evaluate $\int \frac{1}{\sqrt{x^2+x-1}}dx$?
Our professor has given us this integral to think about and I've no idea what to do to solve it. I understand there's probably a substitution somewhere but i can't figure out how. The integral calculators I've used online haven't been helpful enough to explain exactly what to do and why I did it.
First we want to factorise the denominator, in order to set up a convenient substitution: $$\int\frac1{\sqrt{x^2+x+1}}\mathrm{d}x$$ can be rewritten as $$\frac2{\sqrt3}\int\frac1{\sqrt{(\frac2{\sqrt3}x+\frac1{\sqrt3})^2+1}}\mathrm{d}x$$ by completing the square and rescaling the constant term in the denominator. Looking at this integral, one can wishfully think "wouldnt it be nice if the messy squared term was a sinh expression? then the denominator would be $\sqrt{\cosh^2(u)}$, which would cancel very nicely." This leads to the substitution $\sinh u=\frac2{\sqrt3}x+\frac1{\sqrt3}$. Changing the variables in this way turns the integral into $$\int\frac{\cosh u}{\sqrt{sinh^2u+1}}\mathrm{d}u,$$ which is just the integral $\int1\mathrm{d}u$ due to the identity $\cosh^2u=1+\sinh^2u$. The antiderivative is therefore $u=\mathrm{arsinh}(\frac2{\sqrt3}x+\frac1{\sqrt3})$.