# U-substitution problem

1. Dec 8, 2015

### Mr Davis 97

1. The problem statement, all variables and given/known data

$\displaystyle \int \frac{x}{\sqrt{x + 1}}dx$

2. Relevant equations

3. The attempt at a solution

First, I let $u = x + 1$. Then $du = dx$ and $x = u - 1$.

So $\displaystyle \int \frac{u - 1}{\sqrt{u}}du = \int u^{\frac{1}{2}} - u^{-\frac{1}{2}}du = \frac{2}{3}u^{\frac{3}{4}} - 2u^{\frac{1}{2}} + C = \frac{2}{3}u\sqrt{u} - 2u + C = \frac{2}{3}u(\sqrt{u} - 3) + C = \frac{2}{3}(x + 1)(\sqrt{x + 1} - 3) + C$

However, the answer book says that the correct answer is $\displaystyle \frac{2}{3}(x - 1)\sqrt{x + 1} + C$

What am I doing wrong?

2. Dec 8, 2015

### Samy_A

Check what $\int u^{p}du$ is for $p \neq -1$. You made a mistake when computing one of the integrals.

Also, in the above step something went wrong.

Hmm, I find a slightly different result than what your book says ...

Last edited: Dec 8, 2015
3. Dec 8, 2015

### LCKurtz

You don't mean $u^{\frac{3}{4}}$there.
and you don't mean $2u$ there.

Fix that, and with a little care you will find the text answer is correct.

Last edited: Dec 8, 2015
4. Dec 8, 2015

### Ray Vickson

The book's answer is incorrect; differentiate it wrt x to see that it fails to give back the integrand $x/\sqrt{x+1}$.

5. Dec 9, 2015

### LCKurtz

Yes. I made the same silly error the book did. The text answer should have been$$\frac{2}{3}(x - 2)\sqrt{x + 1} + C$$