How do I evaluate this integral: $ \ \int (x^2-1)/ \ln(x) \ \ dx \ \ $?

Question

This is the integral that I have to evaluate:

How do I evaluate the indefinite integral? $$ \int \ \frac{x^2 - 1}{\ln x} \ \ dx \ \ . $$

I was just looking up for some 'tough integrals' to practice on youtube and then this video shows up:

The Source

This guy allegedly uses the Feynman integral trick and I mostly understand what he is writing.

But is there a way the definite integral can be done without using the trick? (as tricks are generally shortcuts.)

It seems extremely hard to do without the trick. I could barely scratch the surface(maybe).

Answer 1

$\int \frac{x^2-1}{\ln x} dx.$ $e^u=x. e^udu=dx$

$\int \frac{e^{2u}-1}{u}e^u du=\int \frac{3e^{3u}du}{3u}-\int \frac{e^udu}{u}$

$v=3u. dv=3du$

$\int \frac{e^vdv}{v}-\int \frac{e^udu}{u}=C$

I think that's it.

Answer 2

$$I=\int \ \frac{x^2 - 1}{\log( x)} \, dx $$ $$x=e^t \quad \implies \quad I=\int \frac{e^t(e^{2t}-1)} t \,dt$$ $$I=\int \frac{e^{3t}}{3t}\,d(3t)-\int \frac{e^{t}}{t}\,dt$$ $$\int\frac{e^{t}}{t}\,dt=\text{Ei}(t)$$ where appears the exponential integral function.