# Homework Help: Integral of a fraction

1. Jun 29, 2008

### aquitaine

2. Relevant equations

$$\int\sqrt{x^{3}}+1/\sqrt{x}+1$$

3. The attempt at a solution

I tried substitution, but that wouldn't work. I tried integration by parts, but I must not have done it properly since my answer was several factors of x off. Am I missing something?

2. Jun 29, 2008

### rocomath

Use parenthesis!!! I'm assuming your problem is:

$$\int\frac{\sqrt{x^3} +1}{\sqrt x+1}dx$$

So, where is your work? Thanks :)

3. Jun 29, 2008

### dirk_mec1

Now it isn't clear what your integral is.

Is it the integral below? Is het root taken only of x^3 and x or the whole term?

$$\int \frac{ \sqrt{x^3}+1}{\sqrt{x}+1} \mbox{d}x$$

4. Jun 29, 2008

### RTW69

Assuming rocomath is correct in formulating the problem, try long division before integration

5. Jun 29, 2008

### aquitaine

Right. Here's my work:

u=x$$^{3/2}$$+ 1 du= $$\frac{3}{2}$$x$$^{1/2}$$
dv = x$$^{1/2}$$ + 1 v= 2x$$^{3/2}$$ + x

After that I decided to check the answer before proceeding further to see if I was on the right track, which I wasn't. The books answer was (1/2)x^2 - (3/2)x$$\sqrt{x}$$ + x + C, and I'm at a loss as to how that happened.

6. Jun 29, 2008

### rocomath

I let $$x=u^2 \rightarrow dx=2udu$$

Then I used polynomial division.

Last edited: Jun 29, 2008
7. Jun 30, 2008

### RTW69

divide sqrt(x)+1 into x*sqrt(x)+1, the integration then becomes very straight forward

8. Jun 30, 2008

### dynamicsolo

Quite so, or factor $$x^{3/2} + 1$$ as a sum of two cubes...

9. Jun 30, 2008

### rocomath

Oh! Very nice, didn't even notice that.

10. Jun 30, 2008

thanks guys!