Integration by change of variable

**doctorgk** · February 29th 2008, 04:47 PM

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

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 $x+1$

thanks

**TwistedOne151** · February 29th 2008, 05:36 PM

make the u substitution $u=\sqrt{x+1}$ . Then $x=u^2-1$ , and $dx=2u\,du$ . 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.

**galactus** · February 29th 2008, 05:40 PM

Your sub will work fine.

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

Make the subs and you get:

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

Now integrate that. Easier, huh?.

**polymerase** · February 29th 2008, 05:46 PM

Originally Posted by doctorgk

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

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 $x+1$

thanks

You can also make the substitution $u^2=x+1$ , you would end up integrating $-2\int u+1\:du$

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