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I faced two similar integrals today.

They are

$$\int\frac{1}{x^2(x^{2009}+1)^{ \frac{2008}{2009}}}dx$$

and $$\int\frac{1}{x^2(x^{2009}+1)^{ \frac{1}{2009}}}dx$$

No trigonometric substitution is working here.I tried almost all.What to do?

Moreover binomial expansion was perhaps possible but that would be really very ugly...

Edit: Oh and yes another one too which is nearly similar but I could'nt solve.

$$\int\frac{1}{{e^x}({e}^{2008x}+1)^{ \frac{2007}{2008}}}dx$$

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    The answer to the first one is simply $-(1+x^{2009})^{1/2009}/x$. For the second one, you will get a similar thing if the $2007/2009$ is changed to $2007/2008$ (or some other change). Otherwise you will get an answer in terms of a Hypergeometric function. – mickep Feb 09 '16 at 04:39
  • Wait a second.How did you get the first answer!Tell me in steps please! –  Feb 09 '16 at 04:41
  • As in the solution of @juantheron... – mickep Feb 09 '16 at 04:49
  • Eh!Well thanks for the reply :-) @mickep –  Feb 09 '16 at 04:50

1 Answers1

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Let $$I = \int\frac{1}{x^2(x^{2009}+1)^{\frac{2008}{2008}}}dx = \int\frac{1}{x^2\cdot x^{2008}(1+x^{-2009})^{\frac{2008}{2009}}}dx$$

Now Put $1+x^{-2009} = t\;,$ Then $\displaystyle x^{-2010}dx = -\frac{1}{2009}dt$

So $$I = -\frac{1}{2009}\int t^{-\frac{2008}{2009}}dt$$

For the second one:: $$I = \int\frac{1}{e^x(e^{2008x}+1)^{\frac{2007}{2008}}}dx\;,$$ Put $e^x=t\;,$ and $e^xdx=dt$

So we get $$I = \int\frac{1}{t^2(t^{2008}+1)^{\frac{2007}{2008}}}dt$$

juantheron
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