# Integration by change of variable

• Feb 29th 2008, 04:47 PM
doctorgk
Integration 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 PM
TwistedOne151
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 PM
galactus

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 PM
polymerase
Quote:

Originally Posted by doctorgk
$\displaystyle \int \frac{-x}{(x+1)-\sqrt (x+1)} dx$

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

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