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I was looking around, and stumbled accross the fact that

$$ \int \frac{\mathrm{d}u}{(1+\sqrt{u})(1+u)^2\,} = \frac{1}{2} \kappa \left( 1 - \frac{1}{\sqrt{u}\,} \right) - \frac{1}{4} \log\left( 1 + 2\kappa \right) $$ Where $\kappa = \sqrt{u}/(1+u)$

I know I have seen this integral before, but can not forthe life of me remember where. Are there any clever substitutions one can make, because partial fractions makes this a nightmare.

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Try $1+\sqrt{u} = t$ it will give easier transformation. If you can't then leave a message I'll do completely.

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  • $\begingroup$ You still end up with three integrals that magically end up canceling each other out at the end of the computation. Is there really nothing simpler than partial fractions? $\endgroup$ May 9, 2013 at 7:15
  • $\begingroup$ actually i don't know how to write mathematical expression here that's why i'm giving hint.ok i'll try my way hope u can understand. $\endgroup$ May 9, 2013 at 7:29

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