Integration of p(1+p2)√1+p2p\over {(1+p^2) \sqrt{1+p^2}}

Find

∫p(1+p2)√1+p2dp\int {p\over {(1+p^2) \sqrt{1+p^2}}} dp

My try

Let u=1+p2u=1+p^2

Then p=√u−1p=\sqrt{u-1}

And dp=12√u−1dudp={1\over {2\sqrt{u-1}}} du

So

∫p(1+p2)√1+p2dp=12∫u−32du\int {p\over {(1+p^2) \sqrt{1+p^2}}} dp={1\over 2}\int {u}^{-3\over 2} du

=−u−12+c=-u^{-1\over 2}+c

=−(1+p2)−12+c=-{(1+p^2)}^{-1\over 2}+c

=−1√1+p2+c={-1\over {\sqrt{1+p^2}}}+c

True?

Thanks.

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Everything is correct.
– JeanMarie
2 days ago

  

 

Thank you so much.
– Dima
2 days ago

  

 

Do not forget to add the dudu
– Claude Leibovici
yesterday

  

 

@ClaudeLeibovici thanks, I added it.
– Dima
yesterday

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