$\displaystyle \int \frac{-x}{(x+1)-\sqrt (x+1)} dx$

Kind of lost, I tried letting u=x+1, then follow through, but my answer does not match up with the right answer

the square root goes over the entire quantity $\displaystyle x+1$

thanks

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- Feb 29th 2008, 04:47 PMdoctorgkIntegration by change of variable
$\displaystyle \int \frac{-x}{(x+1)-\sqrt (x+1)} dx$

Kind of lost, I tried letting u=x+1, then follow through, but my answer does not match up with the right answer

the square root goes over the entire quantity $\displaystyle x+1$

thanks - Feb 29th 2008, 05:36 PMTwistedOne151
make the u substitution $\displaystyle u=\sqrt{x+1}$. Then $\displaystyle x=u^2-1$, and $\displaystyle dx=2u\,du$. When you make this subtitution, factor the numerator and the denominator (they're both polynomials) and cancel common factors. The result is easily integrable.

--Kevin C. - Feb 29th 2008, 05:40 PMgalactus
Your sub will work fine.

Let $\displaystyle u=x+1, \;\ du=dx, \;\ u-1=x$

Make the subs and you get:

$\displaystyle \frac{1-u}{u-\sqrt{u}}=\frac{-1}{\sqrt{u}}-1$

Now integrate that. Easier, huh?. - Feb 29th 2008, 05:46 PMpolymerase