What is the trick to solving the integral of x/sqrt(x + 2)?

[James Brady]

\int\frac{x}{\sqrt{x + 2}}dx

We are still using substation as our method of solving integrals. I've rationalized the denominator, but that doesn't seem to help a whole lot. Any value for u I've picked so far hasn't worked. I've looked up the solution online, and I know it's not a trig integral. Any small hint would help.

 

[CompuChip]

Once you recognise that
$\frac{1}{\sqrt{x + 2}} \propto \frac{d}{dx} \sqrt{x + 2}$
you could try integration by parts.

 

[LCKurtz]

Or you could try $x+2=u^2$.

 

[ehild]

Or u=x+2

ehild

 

[vanhees71]

...or you use a little trick:

<br /> \int \mathrm{d} x \frac{x}{\sqrt{x+2}}=\int \mathrm{d} x \frac{x+2-2}{\sqrt{x+2}} = \int \mathrm{d} x \left [(x+2)^{1/2}-2 (x+2)^{-1/2} \right ]=\frac{2}{3} (x+2)^{3/2} - 4 (x+2)^{1/2}+\text{const}.

 

[James Brady]

...or you use a little trick:

<br /> \int \mathrm{d} x \frac{x}{\sqrt{x+2}}=\int \mathrm{d} x \frac{x+2-2}{\sqrt{x+2}} = \int \mathrm{d} x \left [(x+2)^{1/2}-2 (x+2)^{-1/2} \right ]=\frac{2}{3} (x+2)^{3/2} - 4 (x+2)^{1/2}+\text{const}.

Wow, sweet moves bro... I like that trick. Thanks a million.