# Integrating Natural Log

1. Jul 6, 2007

### benedwards2020

I'm having real problems integrating a natural log. The problem I have been set is (where S = integration sign)

S (x - 2) Ln(3x)

I'm trying to use the integration by parts rule but keep getting the wrong answer and I think it might be to do with the natural log. I have used

f(x) = Ln(3x) f'(x) = 1/x g'(x) = (x - 2) and g(x) = 1/2x^2 - 2x

Using the integration by parts formula, I end up with

Ln(3x) * x^2/2 - 2x + S1/2x - 2x

The answer is wrong..... Can someone help me out here.... Where am I going wrong?

2. Jul 6, 2007

### mjsd

observe that
$$\int \log(x) dx= -x +x \log(x)$$ from integration by parts and
$$\int x\log(x) dx=-x^2/4+ x^2 \log(x)/2$$ also from integration by parts

just check you workings again.

3. Jul 6, 2007

### Dick

Use parentheses where needed! Like what is 1/2x? Is it (1/2)*x or 1/(2*x)??? But check again what the product f'(x)*g(x) is.

4. Jul 6, 2007

### D H

Staff Emeritus
Watch your parentheses and review the integration by parts rule. Hint: You have a sign error.

5. Jul 6, 2007

### benedwards2020

Am I right in assuming that I am to use the integration by parts method and the values that I have chosen for f(x), f'(x), g'(x) & g(x) are correct?

Apologies for the parenthesis error.... I'm trying to learn how to use the equation package....

6. Jul 6, 2007

### benedwards2020

Ok, here's a go at using Latex.... Hope this makes what I'm doing a bit clearer...

Right, From the initial problem

$$\int (x - 2) \ln(3x)$$ I have used

$$f(x) = \ln(3x) \ f'(x) = \frac{1}{x} \ g(x) = \frac{1}{2}x^2 - 2x \ g'(x) = (x - 2)$$

By using the integration by parts formula, I have

$$\ln(3x) \times \frac{1}{2}x^2 -2x - \int \frac{1}{x} \times \frac{1}{2} x^2-2x$$

Am I right in saying this up to now?

Last edited: Jul 6, 2007
7. Jul 6, 2007

### cristo

Staff Emeritus
With the brackets in, it's correct.

Last edited: Jul 6, 2007
8. Jul 6, 2007

### benedwards2020

So we have

$$\int \frac{1}{x} \left[ \frac{1}{2}x^2 - 2x \right] dx \longrightarrow \frac{1}{4}x^2 - 2x$$

to give us

$$\ln(3x) \left[ \frac{1}{2}x^2 - 2x \right] - \left[ \frac{1}{4}x^2 - 2x \right]$$

Does this look ok? I think that maybe this is where I'm going wrong... However, I suspect my brackets were in the wrong place....

Last edited: Jul 6, 2007
9. Jul 6, 2007

### NateTG

Looks good to me.

10. Jul 7, 2007