# Simple integration question

1. Nov 30, 2007

### 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.

Last edited: Nov 30, 2007
2. Nov 30, 2007

### 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.

3. Dec 3, 2007

### 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.

4. Dec 3, 2007

### 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.

5. Dec 3, 2007

### Kurdt

Staff Emeritus
There appears to be a mistake in the first substitution as Dick has pointed out. If thats printed in a text book then it appears to be a misprint.