Solving U-Substitution Problem: Integrate x/sqrt(x+1)dx

[Mr Davis 97]

Homework Statement

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

Homework Equations

The Attempt at a Solution

[/B]
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?

 

[Samy_A]

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

$\frac{2}{3}u^{\frac{3}{4}} - 2u^{\frac{1}{2}} + C = \frac{2}{3}u\sqrt{u} - 2u + C$

Also, in the above step something went wrong.

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

So $\displaystyle \int \frac{u - 1}{\sqrt{u}}du = \int u^{\frac{1}{2}} - u^{-\frac{1}{2}}du = \frac{2}{3}u^{\frac{3}{4}}$

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

$- 2u^{\frac{1}{2}} + C = \frac{2}{3}u\sqrt{u} - 2u + C$

and you don't mean $2u$ there.

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

The Attempt at a Solution

[/B]
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?

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

 

[LCKurtz]

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?

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

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