# How to find $\int{\left(\frac{\sqrt{x+1}}{x-1}\right)^x}dx$?

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

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You are sure this has an elementary antiderivative? Where is this function from? – martini Apr 27 '12 at 6:59
Wolfram Alpha can't find an elementary antiderivative (and neither can I!) – user29743 Apr 27 '12 at 7:01
This looks evil! – Tomarinator Apr 27 '12 at 7:23
@martini No. That's why I've asked. The function comes from my imagination… – Garmen1778 Apr 27 '12 at 16:29
@Garmen1778 The problem with our imagination is that it can create problems which noone can solve ;) – N. S. May 3 '12 at 20:00

You can do... $$\int f(x)^x\; dx=\int e^{x\ln f(x)}\; dx=\int e^{\alpha(x)}\; dx$$ where $f(x)=\frac{\sqrt{x+1}}{x-1}$
that logarithm isn't a neperian logarithm, expressed with $\ln f(x)$? – Garmen1778 May 8 '12 at 21:09
Ups, sorry, I mean natural logarithm, expressed by $\ln f(x)$ or $\log_e f(x)$ – Garmen1778 May 9 '12 at 5:37