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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

Zuo
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  • Just as a hint, Chebyshev's theorem here(https://www.encyclopediaofmath.org/index.php/Chebyshev_theorem_on_the_integration_of_binomial_differentials), 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
  • For the case, $n=1$ it is obvious. For the case $n=2$, it can be done using Euler substitution. Generally, in the kind of cases one mostly use trigonometric function to substitute something like $t=tgz$. But it would be helpful when you have $\sqrt{t^{2}+1}$, rather than $\sqrt{t^{2}-1}$. So Euler substitution, can be used. – kolobokish Dec 17 '16 at 14:07

3 Answers3

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

kingW3
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It is equal to $x{(1+x^n)}^{-1/n}+C$.

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

lab bhattacharjee
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