- #1

- 347

- 2

## Homework Statement

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

[tex]

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

[/tex]

## Homework Equations

?

## 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.

[itex]

\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

[/itex]

And then equated the coefficients:

For X^2: [itex]4=2A+B[/itex]

For X^1: [itex]6=3B+C[/itex]

For X^0: [itex]-1=3C-A[/itex]

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

[itex]

6=3(2-6C)+C\\

6=6-11C \\

0=-11C \\

∴ C=0

[/itex]

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

Which then leads to:

[itex]

\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|}

[/itex]

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