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

Question

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 $\dfrac{u}{v}$.

You are sure this has an elementary antiderivative? Where is this function from? — martini, Apr 27 '12 at 06:59
Wolfram Alpha can't find an elementary antiderivative (and neither can I!) — , Apr 27 '12 at 07:01
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. — Parsa, Apr 27 '12 at 08:01
@martini No. That's why I've asked. The function comes from my imagination… — Garmen1778, Apr 27 '12 at 16:29
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. — graveolensa, Apr 27 '12 at 17:18
@Garmen1778 The problem with our imagination is that it can create problems which noone can solve ;) — N. S., May 03 '12 at 20:00
@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) — yiyi, May 04 '12 at 16:12
I think it is the power of x which is what gets in the way with finding an explicit formula for this integral. — Comic Book Guy, May 05 '12 at 13:20
I don't think that it's even possible to find $\int{x^x}dx$ ... — Théophile, May 07 '12 at 18:54
For $\int x^x dx$ check http://math.stackexchange.com/questions/141347/how-to-solve-int-xxdx (But this one here looks complicated) — Kirthi Raman, May 08 '12 at 00:37
@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. — Tomarinator, May 08 '12 at 13:33

diofanto · Accepted Answer · 2012-05-09T21:10:35.147

1

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

edited May 09 '12 at 21:10

answered May 08 '12 at 12:52

diofanto

122

-1 Yes, you can, but does it lead you anywhere? – Sasha May 08 '12 at 14:46
Mmm... It was a simple idea. (¬¬) – diofanto May 08 '12 at 15:17
that logarithm isn't a neperian logarithm, expressed with $\ln f(x)$? – Garmen1778 May 08 '12 at 21:09
@Garmen1778 I have tried googling "neperian logarithms" couldn't find that term. Could you tell me what it is? – yiyi May 09 '12 at 02:45
Ups, sorry, I mean natural logarithm, expressed by $\ln f(x)$ or $\log_e f(x)$ – Garmen1778 May 09 '12 at 05:37
@Garmen1778 is that how its said in another language? – yiyi May 09 '12 at 10:37
@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_logarithm – Garmen1778 May 09 '12 at 14:18
@Garmen1778 what! I searched wikipedia.org. Also, which language are you talking about? – yiyi May 09 '12 at 15:26
@MaoYiyi Spanish. – Garmen1778 May 09 '12 at 16:18
The cuestion about natural o decimal logarithm.. Look the exponential!!! (Edited notation) – diofanto May 09 '12 at 21:13

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

1 Answers1