# Integration using Partial Fractions

1. Feb 27, 2006

### Hootenanny

Staff Emeritus
I need to find the following intergral:
$$\int_{0}^{1} \frac{28x^2}{(2x+1)(3-x)} \;\; dx$$
So I split it into partial fractions thus:
$$\frac{2}{2x+1} + \frac{36}{3-x} - 14$$
Then integrated:
$$\int_{0}^{1} \frac{2}{2x+1} + \frac{36}{3-x} - 14 \;\; dx$$
$$= \left[ \ln\left| 2x+1 \right| + 12\ln\left| 2-x \right| - 14x \right]_{0}^{1}$$
But this isn't going to give me the correct answer which is quoted as:
$$37\ln 3 - 36\ln 2 - 14$$
Can anybody see where I've gone wrong? Thank's

2. Feb 27, 2006

### Hootenanny

Staff Emeritus
Sorry just a correction in my answer:
$$= \left[ \ln\left| 2x+1 \right| + 12\ln\left| 3-x \right| - 14x \right]_{0}^{1}$$

3. Feb 27, 2006

### Physics Monkey

Your partial fraction expansion is ok. However, you integrated $$\frac{36}{3 - x}$$ incorrectly.

4. Feb 27, 2006

### Hootenanny

Staff Emeritus
Ahh yes. Thank-you, I realise what I've done now.

5. Feb 27, 2006

### Hootenanny

Staff Emeritus
Is there any standard intergral in the form of $\int \frac{a}{b+x} = ...$ because at the moment I'm not sure what to do. If anybody could guide me through it?

6. Feb 27, 2006

### Kamataat

I suppose you can use $$\int\frac{a}{b\pm x}dx=\pm a\ln|x\pm b|$$

- Kamataat

7. Feb 27, 2006

### Hootenanny

Staff Emeritus
Thank you very much.

8. Dec 29, 2007

### jellybeanzgir

hi there, I was wondering if you could possibly help me with an integration problem

$$\int_{1/4}^{-1/4} \frac{1}{(1-4x^2)} \;\; dx$$

i managed to split it into partial fractions and hence split up the integration as follows

$$\frac{1}{2} \int_{1/4}^{-1/4} \frac{1}{2x+1} - \frac{1}{2x-1} \;\;dx$$

after this i integrated it in the following manner but im unsure as to whether it is right, and if so, how to continue it.

$$[\frac{1}{2} (\frac{1}{2}\ln(2x+1)-\frac{1}{2}\ln(2x-1))]$$still between the same two limits

Can anyone offer a hand?

9. Dec 29, 2007

### jellybeanzgir

if my integration was correct please continue reading this post, if not then it doesn't matter!!

I broke down that inegraton to read

$$[\frac{1}{2} \ln\frac{2x+1}{2x-1}]$$ limits $$\pm\frac{1}{4}$$

i then simplified this down to the following (including limit values)

$$\frac{1}{2} \ln\frac{1.5}{-0.5} - \frac{1}{2}\ln\frac{0.5}{-1.5}$$

Using the rules of logs i calculated this to be

$$\ln\frac{-3}{\frac{-1}{3}}$$

Giving a final answer of $$\ln9$$

Does anyone know if this is the correct answer or have i gone about it in the wrong way?

10. Dec 29, 2007

### Rainbow Child

The correct answer is

$$\frac{1}{4} \left(\ln|2x+1|-\ln|-2x+1|\right)$$

you need the absolute values in $$\ln(x)$$ in order to be well defined.

See Kamataat's post $$\int\frac{a}{b\pm x}dx=\pm a\ln|x\pm b|$$

11. Dec 29, 2007

### rocomath

well you have $$\frac{1}{1-4x^2}$$

so it should become

$$\frac{1}{(1+2x)(1-2x)}$$

if you want to write it as (2x+1)(2x-1), then you have to factor out a negative

12. Dec 29, 2007

### rocomath

you have it the other way around, i even checked it on my calculator and my answer matches:

after integrating it should become

$$\frac{1}{4}\left[\ln\left|\frac{1-2x}{1+2x}\right|\right]_{.25}^{-.25}$$

13. Dec 29, 2007

thanks