Cómo encontrar a $\int{\left(\frac{\sqrt{x+1}}{x-1}\right)^x}dx$?

Question

He tratado de encontrar $$\int{\biggl(\dfrac{\sqrt{x+1}}{x-1}\biggr)^x}dx$$ but I don't know how to do it, because it combines $u^x$ and $\dfrac{u}{v}$.

Answer 1

Usted puede hacer... $$\int f(x)^x\; dx=\int e^{x\ln f(x)}\; dx=\int e^{\alpha(x)}\; dx$$ donde $f(x)=\frac{\sqrt{x+1}}{x-1}$