Integrating \int \frac{x}{sqrt{x+3}}dx: Simple Integration Methods Explained

[John O' Meara]

Integrate \int \frac{x}{sqrt{x+3}}dx \\
(i) I can integrate this by letting U^2=x+3 => u=\sqrt{x+3} \ \mbox{ so } \ x=u^2-3 \ \mbox{and } \ dx=2udu \ \mbox{then the integral = } \ \int\frac{(u^2-3)2u}{u}du \\
\ =2(\frac{u^3}{3}-3u) + C \\ \ = \ \frac{2}{3}(u^2-9) + C \ = \ \frac{2}{3}(x-6)\sqrt{x+3} +C \\
Or (ii) I can integrate above if I let u=x+3 => \ \frac{du}{dx}=1 \ \mbox{ therefore } \ du = dx \ \mbox{ and } \ x=u-3 \ \\. \mbox{ then } = \ \int\frac{u-3}{u^{\frac{1}{2}} du \ = \ \int u^{\frac{1}{2}}du \ -\ 3 \int u^{\frac{-1}{2}} du \ = \ \frac{2}{3} u^{\frac{3}{2}} \ - \ 6u^{\frac{1}{2}} +C \ \\.\ = \ \frac{2}{3}(x+3)^{\frac{3}{2}} \ -6(x+3)^{\frac{1}{2}} + C\\. Which is correct (i) or (ii), and why? Thanks alot.

 

[Dick]

The substitution is fine on the first one. But how does 2(u^3/3-3u) turn into (2/3)*(u^2-9)? The second one is fine.

 

[John O' Meara]

So you are telling me that both substitutions are valid? That is answer (i) and (ii) are equal? It is a mis-print.

 

[coomast]

Indeed, both substitutions are valid. Sometimes there are several ways to come to a solution in doing integrals. One is often more preferable over the other due the length of the algebra involved. Practice is the only way to recognize the appropriate one. In your case both are equally usable.

The first solution (with a minor typo) can be rewritten as:

\frac{2}{3}\sqrt{x+3}\cdot (x-6)+C=\frac{2}{3}\sqrt{x+3}\cdot (x-9+3)+C

Expanding this gives you the second solution, which is therefore proven to be the same.

Sometimes you need to adjust the integration constant, to show that two solutions are indeed the same, but not here.

 

[Kurdt]

There appears to be a mistake in the first substitution as Dick has pointed out. If that's printed in a textbook then it appears to be a misprint.