# Compute $\int\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx$

How to compute this integral:$$\int\frac{1}{(1+x^n)\sqrt[n]{1+x^n}}dx$$I tried to make$$\ t^n = 1+x^n$$ But I got a more complicated formula$$\int\frac{1}{t^{n+1}}\frac{t^{n-1}}{{(t^n-1)}^\frac{n-1}{n}}dt$$then I can not go on

• Just as a hint, Chebyshev's theorem here(encyclopediaofmath.org/index.php/…), states that your integral has solution in elementary functions, only in these case 1)$n=1$ 2) $n=2$ 3)$n=3$ – kolobokish Dec 17 '16 at 12:55
• Ah, I'm sorry, i didn't notice the $dx$, I thought it was $dt$ – kolobokish Dec 17 '16 at 12:58
• No. I was right. There should be $dt$ in the last integral. – kolobokish Dec 17 '16 at 13:01
• Sorry for not typing the whole thing, but you can do the following, for case $n=2$, take one of Euler substitution(look up in any calculus book, or in wikipedia), and than if necessary split any rational function you'll get into seperate simplier fractions. (Sometimes it is called Lagrange method of rational integrals). It may help, however I'm sorry, I tried a little, and then let it. – kolobokish Dec 17 '16 at 13:33
• @kolobokish:Thank you for your reply.According to the Chebyshev's theorem, it can be expressed as an elementary function.But I have not yet calculated – Zuo Dec 17 '16 at 13:37

## 3 Answers

$$t=(1+x^n)^{-1/n}\\dt=-\frac{x^{n-1}}{(1+x^n)^{-(n+1)/n}}dx\\t^{-n}=1+x^n\\x^{n-1}=(t^{-n}-1)^{(n-1)/n}\\x^{n-1}=\frac{(1-t^n)^{(n-1)/n}}{t^{n-1}}\\\int-\frac{t^{n-1}}{(1-t^n)^{(n-1)/n}}dt\\(1-t^n)^{1/n}=c\\-\frac{t^{n-1}}{(1-t^n)^{(n-1)/n}}dt=dc\\\int dc=c+C=(1-t^n)^{1/n}=(1-(1+x^n)^{-1})^{1/n}=(1-\frac{1}{1+x^n})^{1/n}=(\frac{x^n}{1+x^n})^{1/n}+C=x(1+x^n)^{-1/n}+C$$

• Thank you very much for your help,I get it! – Zuo Dec 17 '16 at 17:04

It is equal to $x{(1+x^n)}^{-1/n}+C$.

• Could you explain how to get this formula? – Zuo Dec 17 '16 at 13:42
• Actually I take n=0,1,2 and get the result and then induct the final one. – Jason Dec 17 '16 at 14:16

HINT:

Choose $x=1/y$

$\implies dx=-\dfrac{dy}{y^2}$

and $\dfrac1{(1+x^n)^{1+1/n}}=\dfrac{y^{n+1}}{(y^n+1)^{1+1/n}}$

Now set $y^n+1=u$

Or directly, $u=y^n+1=\dfrac{x^n}{1+x^n}$