# Integration by Partial Fractions

1. May 20, 2014

1. The problem statement, all variables and given/known data

Find the indefinite integral of the below, using partial fractions.

$$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}$$

2. Relevant equations
?

3. The attempt at a solution
First I want to say there is probably a much easier and quicker way to get around certain things I have done but I have just done it the only way I could see (I always seem to go the long way around). So if I am correct then I would really appreciate some tips on how to do it quicker (for in exams) and also if I am incorrect to point out my mistakes. Thanks :)

first I set up the partial fractions.
$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{A}{x+3}+\frac{Bx+C}{2x^2-1} \\ \rightarrow \,\,\, 4x^2+6x-1=A(2x^2-1)+(Bx+C)(x+3) \\ 4x^2+6x-1=2Ax^2-A+Bx^2+3Bx+Cx+3C$

And then equated the coefficients:
For X^2: $4=2A+B$
For X^1: $6=3B+C$
For X^0: $-1=3C-A$

Then what I did what multiply the third equation above by 2 to get $-2=6C-2A$ and then added it to the first equation to get $2=6C+B$ and solved for B and substituted it into the second equation which resulted in:

$6=3(2-6C)+C\\ 6=6-11C \\ 0=-11C \\ ∴ C=0$

And then knowing C=0 I found A=1 and then that B=2 .

$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{1}{x+3}+\frac{2x}{2x^2-1} \\ \frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{1}{x+3}+\frac{2x}{2x^2-1}\\ \int \frac{4x^2+6x-1}{(x+3)(2x^2-1)} \,\, dx = \ln{|x+3|} + \frac{1}{2} \ln{|2x^2-1|}$

I think I may have done something wrong. I had to use an integral calculator online to find the integral of $\frac{2x}{2x^2-1}$ as I had no idea how to do it. Which kind of defeats the point.

2. May 20, 2014

### Saitama

To evaluate the last integral, you can use the substitution $2x^2-1=t$.

3. May 20, 2014

### Panphobia

Its all right, except you need to add a + C. But by the way the integral of 2x/(2x^2-1) can be easily solved using u substitution.

If u = 2x^2 -1 then du = 4x dx so du/2 = 2x dx

(1/2)*(1/u)

and the integral of this is

ln|u|/2

which is

ln|2x^2-1|/2

Last edited: May 20, 2014
4. May 20, 2014

### Curious3141

You're missing the constant of integration, which is very important in an indefinite integral.

The partial fraction decomposition is tedious. If the denominator was easily factored into linear factors, Heaviside cover up rule would help more. As it stands, the cover up rule helps to find A quite quickly, but you get bogged down with B and C, and you have to manipulate radicals. Pointless. I think your method is fine.

5. May 20, 2014

Ah yes, sorry I forgot as it was just rough workings copied up into latex.

I am quite poor when it comes to integration by substitution for some reason. I find integration by parts easier as its a set formula but I do realise that integration by substitution is sometimes a lot easier and quicker so I need to get more familiar with it. Any tips on u substitution?

I can usually find what to substitute, and then get to a du=... but its the following bits that confuse me a little.

6. May 20, 2014

### scurty

I try to use u-substitution if I see a power that is one higher in the denominator than in the numerator. A more general approach would to be to look for derivatives in the numerator from what is found in the denominator. I always look for u-substitution first because they tend to be the easiest when dealing with fractions. Partial fractions and trig substitution, while fun to compute, are more time consuming.

It's hard for me to explain because I've done so many integrals that knowing what method to use is more intuitive to me than anything else. It comes easier the more you practice integrals. A general rule to always follow for u substitution is to notice candidates for u and du.