# Quick Question

1. Sep 10, 2005

### laker88116

$$\int \frac {1}{x\sqrt{4x+1}}dx$$

Here's what I have done so far on this problem

I let $$u= \sqrt{4x+1}$$, so then $$u^2=4x+1$$, $$du= \frac {2dx}{u}$$ and $$x= \frac {u^2-1}{4}$$

Substituting, I get $$\int \frac {1}{(\frac{u^2-1}{4})u}du$$

Then moving stuff around, I get $$4 \int \frac {du}{u(u+1)(u-1)}$$

I know I have to use partial fractions. But I am not sure where to start. Any help is appreciated.

2. Sep 10, 2005

### amcavoy

$$\frac{1}{u(u+1)(u-1)}=\frac{A}{u}+\frac{B}{u+1}+\frac{C}{u-1}$$

3. Sep 10, 2005

### laker88116

I know that, but then what do I do, I have to get a common denominator, which is $$u(u+1)(u-1)$$, but then what?

4. Sep 10, 2005

### amcavoy

Multiply both sides by u(u+1)(u-1) so that the left side is 1. Simplify to get A, B, and C.

5. Sep 10, 2005

### laker88116

Ok, I get A=-1, b=1/2, c=1/2, is that right?

6. Sep 10, 2005

### amcavoy

Looks good.

7. Sep 10, 2005

### laker88116

Ok, I think I got it. Thanks.

8. Sep 11, 2005

### Fermat

Just a little bit missed out. When substituting into the integrand written in terms of x, you forgot to replace the dx with (u/2)du.

$$\mbox{let}\ u= \sqrt{4x+1}\ \mbox{, so then}\ u^2=4x+1\ \mbox{,}\ du= \frac {2dx}{u}\ \rightarrow dx = \frac{u}{2}\ du \mbox{, and}\ x= \frac {u^2-1}{4}$$

Substituting gives $$\int \frac {1}{x\sqrt{4x+1}}dx = \int \frac {1}{(\frac{u^2-1}{4})u}.\frac{u}{2}\ du = \int \frac{2}{u^2 - 1}\ du$$

Now you can do the partial fractions bit.