Is there any trick to calculate $\int\ln(x+\sqrt{x^2-1})dx$ faster? $$\int\ln(x+\sqrt{x^2-1})dx$$
To evaluate the integral we can apply integral by parts (using $u=\ln(x+\sqrt{x^2-1})$ and $dv=dx$). but my question is: is there other approach to evaluate the integral faster? I think there might be something because we can write it as:
$$\ln\left(x+\sqrt{x^2-1}\times\dfrac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\right)=\ln\left(\cfrac1{x-\sqrt{x^2-1}}\right)=-\ln(x-\sqrt{x^2-1})$$
But I don't know how to use this.
Use $x=\cosh t$
$$\int \sinh t \ln(\cosh t + \sinh t) dt = \int t \sinh t dt = t\cosh t - \sinh t$$
then substitute back.
Here is the solution of your integral after you simplified it. I used Symbolab