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$\begingroup$

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?2012-04-27
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    Wolfram Alpha can't find an elementary antiderivative (and neither can I!)2012-04-27
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    This looks evil!2012-04-27
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    If this integral is from 2 to infinity, you can show it converges by using the integral test to make it into a series and then applying the root test. I think that's probably what they want you to do.2012-04-27
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    I agree with @Parsa. I would rewrite as a series.2012-04-27
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    @martini No. That's why I've asked. The function comes from my imagination…2012-04-27
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    I put the function into wolfram alpha to get a series representation, though I couldn't find any sequences on oeis.org that the numerators and denominators correspond to -- the series representation within its radius of convergence is in fact nice in the regard that it has rational numerators and denominators on the coefficients.2012-04-27
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    @Garmen1778 The problem with our imagination is that it can create problems which noone can solve ;)2012-05-03
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    Not evil, just improper (at $x=1$).2012-05-03
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    @Garmen1778 don't worry, if your bounty doesn't solve it, I will put a bounty on it...(checking to see how to do that)2012-05-04
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    I think it is the power of x which is what gets in the way with finding an explicit formula for this integral.2012-05-05
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    I don't think that it's even possible to find $\int{x^x}dx$ ...2012-05-07
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    For $\int x^x dx$ check http://math.stackexchange.com/questions/141347/how-to-solve-int-xxdx (But this one here looks complicated)2012-05-08
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    @KVRaman Both questions are mine.2012-05-08
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    @Garmen1778 there are some functions which ,(they say) cannot be integrated (without mathematica), this one from your imagination might belong to that set of questions.2012-05-08

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You can do... $$\int f(x)^x\; dx=\int e^{x\ln f(x)}\; dx=\int e^{\alpha(x)}\; dx$$ where $f(x)=\frac{\sqrt{x+1}}{x-1}$

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    -1 Yes, you can, but does it lead you anywhere?2012-05-08
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    Mmm... It was a simple idea. (¬¬)2012-05-08
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    that logarithm isn't a neperian logarithm, expressed with $\ln f(x)$?2012-05-08
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    @Garmen1778 I have tried googling "neperian logarithms" couldn't find that term. Could you tell me what it is?2012-05-09
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    Ups, sorry, I mean _natural logarithm_, expressed by $\ln f(x)$ or $\log_e f(x)$2012-05-09
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    @Garmen1778 is that how its said in another language?2012-05-09
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    @MaoYiyi Yes and no. **Yes** because here we say "neperian logarithm"(in our language) because the teachers taught us wrong. And **no** because the real definition of "neperian logarithm" is shown in this page: http://en.wikipedia.org/wiki/Napierian_logarithm2012-05-09
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    @Garmen1778 what! I searched wikipedia.org. Also, which language are you talking about?2012-05-09
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    @MaoYiyi Spanish.2012-05-09
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    The cuestion about natural o decimal logarithm.. **Look** the exponential!!! (Edited notation)2012-05-09