# Thread: Another Partial Fraction Integration

1. ## Another Partial Fraction Integration

Hi
Can you tell me if i am correct here?
thanks yet again
John

Using partial fractions Integrate

$\dfrac{(3x+1)}{(x^2)(2x+4)}$

$\dfrac{(3x+1)}{(x^2)(2x+4)}=\frac{A}{(x^2)}+\frac{ B}{(2x+4)}....equation (1)$

Step 1:

Multiply both sides by $(x^2)(2x+4)$ gives;

$(3x+1)=(2x+4).A +(x^2).B$

Step 2:

Make $x=-2$

$(3(-2)+1)=(2(-2)+4).A +((-2)^2).B$ gives;

$\frac {-5}{4}=B$

Step 3:

Make $x=0$

$(3(0)+1)=(2(0)+4).A +((0)^2).B$ gives;

$\frac {1}{4}=A$

Step 4:

Substitute into equation (1) for A & B

$\dfrac{(3x+1)}{(x^2)(2x+4)}=\frac{1}{4(x^2)}-\frac{5}{4(2x+4)}$

Step 5:

now integrate the separate terms;

$\int \dfrac{(3x+1).dx}{(x^2)(2x+4)}=\int \frac{1.dx}{4(x^2)}-\int \frac{5.dx}{4(2x+4)}$

Step 6:

$=\frac{1}{4}\int \frac{dx}{(x^2)}-\frac{5}{4}\int \frac{dx}{(2x+4)}$

Step 7:

$=\frac{1}{4} ln(x^2) -\frac{5}{4}\frac{ln(2x+4)}{2} +C$

Step 8:

$=\frac{1}{4} ln(x^2) -\frac{5}{8}ln(2x+4) +C$

Step 9:

$=\frac{2}{8} ln(x^2) -\frac{5}{8}ln(2x+4) +C$

Step 10:

$=\frac{-3}{8} [ln(x^2) -ln(2x+4)] +C$

Step 11:

$=\frac{-3}{8} ln[(x^2) -(2x+4)] +C$

2. ## Re: Another Partial Fraction Integration

The degree of the denominator has always to be greater (one degree) then the degree of the numerator, in this case you can write:
$\frac{3x+1}{x^2(2x+4)}=\frac{Ax+B}{x^2}+\frac{C}{2 x+4}$
$\frac{(Ax+B)(2x+4)+Cx^2}{x^2(2x+4)}=\frac{2Ax^2+4A x+2Bx+4B+Cx^2}{x^2(2x+4)}$
$=\frac{(2A+C)x^2+(4A+2B)x+4B}{x^2(2x+4)}$

Therefore:
$4A+2B=3$ and $4B=1 \Leftrightarrow B=\frac{1}{4}$
And so $4A+\frac{1}{2}=3 \Leftrightarrow A=\frac{5}{8}$
And so $2A+C=0 \Leftrightarrow C=\frac{-5}{4}$

3. ## Re: Another Partial Fraction Integration

thanks Siron
I am a bit lost here?

My book does not cover this at all!!

Can you elaborate on "The degree of the denominator has always to be greater (one degree) then the degree of the numerator"

Where is the $\frac{(3x+1)}{x^2(2x+4)}=\frac{Ax+B}{x^2}+\frac{C} {(2x+4)} coming from?$

apologies if the is easy but i am lost
appreciate the help if you have the time to explain it?
John

4. ## Re: Another Partial Fraction Integration

I understand, but it's a little bit long to explain I think so I recommand you to do some research in the book you use. Try to look for some examples etc ... and afterwards try this exercice again .

ok will do
thanks