Partial Fraction Integration

**Simplicity** · May 4th 2008, 01:10 AM

Q: $f(x) \equiv \frac{1}{(x+2)(x^2+4)}$ . Express $f(x)$ in partial fractions.

Show that $\displaystyle\int^2_0 f(x) \, \mathrm{d}x = \frac{(\pi + 2 \ln 2)}{32}$

I can't seem to get the partial fraction right. If possibe, can someone help me with this question for me. Thanks.

.

**Jhevon** · May 4th 2008, 01:16 AM

Originally Posted by Air

Q: $f(x) \equiv \frac{1}{(x+2)(x^2+4)}$ . Express $f(x)$ in partial fractions.

Show that $\displaystyle\int^2_0 f(x) \, \mathrm{d}x = \frac{(\pi + 2 \ln 2)}{32}$

I can't seem to get the partial fraction right. If possibe, can someone help me with this question for me. Thanks.

.

did you start by doing $\frac 1{(x + 2)(x^2 + 4)} = \frac A{x + 2} + \frac {Bx + C}{x^2 + 4}$ ?

**Moo** · May 4th 2008, 01:18 AM

Hello,

$\frac{1}{(x+2)(x^2+4)}=\frac{a}{x+2}+\frac{bx+c}{x ^2+4}$

What did you do for your partial fraction ? (It's always better to point out mistakes than doing the whole thing

)

**Simplicity** · May 4th 2008, 01:22 AM

I was doing $\frac{1}{(x+2)(x^2+4)}=\frac{a}{x+2}+\frac{b}{x^2+ 4}$ hence my mistake.

**o_O** · May 4th 2008, 01:24 AM

Splitting into partial fractions: $\frac{1}{(x+2)\left(x^{2}+4\right)} = \frac{A}{x+2} + \frac{Bx + C}{x^{2} + 4}$

So, multiplying both sides by $(x+2)(x^2 + 4)$ we have:
$1 = A\left(x^{2}+4\right) + (Bx + C)(x + 2)$
$1 = Ax^{2} + 4A + Bx^{2} + (2B+C)x + 2C$ (multiplied it out)
$1 = (A+B)x^{2} + (2B + C)x + 4A + 2C$ (grouped)

If we write the left hand side as a quadratic:
$0x^{2} + 0x + 1 = (A+B)x^{2} + (2B + C)x + (4A + 2C)$

We can equate the coefficients to see that:
$A + B = 0$
$2B + C = 0$
$4A + 2C = 1$

So this should be an easy system to solve for and you should be on your way.

Edit: Ok a little too late lol xD.

**Moo** · May 4th 2008, 01:27 AM

Another way of doing it is :

- firstly, multiply by (x+2), then take x=-2. It'll directly yield A.

- secondly, take x=0, to directly get C.

- thirdly...make your own way to find B

**o_O** · May 4th 2008, 01:29 AM

Originally Posted by Moo

Another way of doing it is :

- firstly, multiply by (x+2), then take x=-2. It'll directly yield A.

- secondly, take x=0, to directly get C.

- thirdly...make your own way to find B

Yes ... That's what I meant ...

**Moo** · May 4th 2008, 01:32 AM

Originally Posted by o_O

Yes ... That's what I meant ...

You multiplied all in once by (x+2)(x²+4) ^^

Actually, I was taught this method for things containing more partial fractions, it was really faster

**Jhevon** · May 4th 2008, 02:27 AM

Originally Posted by Air

I was doing $\frac{1}{(x+2)(x^2+4)}=\frac{a}{x+2}+\frac{b}{x^2+ 4}$ hence my mistake.

yes. with partial fractions, the numerator must be one less degree than the denominator. so if it was $\frac 1{(x + 3)(x^3 + 4x + 2)}$ for instance. your partial fractions would be $\frac A{x + 3} + \frac {Bx^2 + Cx + D}{x^3 + 4x + 2}$

there are other rules for partial fractions. like decomposing $\frac 1{x^2}$ becomes $\frac Ax + \frac B{x^2}$ for instance. make sure you look these rules up

**Gusbob** · May 4th 2008, 04:16 AM

there are other rules for partial fractions. like decomposing $\frac 1{x^2}$ becomes $\frac Ax + B{x^2}$ for instance. make sur eyou look these rules up

Wait a second, is the decompostition of $\frac 1{x^2}=$ $\frac Ax + B{x^2}$ ? Or was that a typo for $\frac {A}{x} + \frac{B}{x^2}$ ?

If it was the first one I failed my test miserably.

**mr fantastic** · May 4th 2008, 04:21 AM

Originally Posted by Gusbob

Wait a second, is the decompostition of $\frac 1{x^2}=$ $\frac Ax + B{x^2}$ ? Or was that a typo for $\frac {A}{x} + \frac{B}{x^2}$ ?

If it was the first one I failed my test miserably.

Don't panic. If the factor in the denominator is x^2 then the corresponding partial fractions are $\frac {A}{x} + \frac{B}{x^2}$ .

**Krizalid** · May 4th 2008, 07:07 AM

Originally Posted by Air

Q: $f(x) \equiv \frac{1}{(x+2)(x^2+4)}$ . Express $f(x)$ in partial fractions.

Show that $\displaystyle\int^2_0 f(x) \, \mathrm{d}x = \frac{(\pi + 2 \ln 2)}{32}$

$\begin{aligned} \frac{1}{(x+2)\left( x^{2}+4 \right)}&=\frac{\left( x^{2}+4 \right)-\left( x^{2}-4 \right)}{8(x+2)\left( x^{2}+4 \right)} \\ & =\frac{1}{8}\left\{ \frac{1}{x+2}-\frac{x-2}{x^{2}+4} \right\} \\ & =\frac{1}{8}\left\{ \frac{1}{x+2}-\frac{x}{x^{2}+4}+\frac{2}{x^{2}+4} \right\}. \end{aligned}$

Integrate.

**Jhevon** · May 4th 2008, 01:42 PM

Originally Posted by Gusbob

Wait a second, is the decompostition of $\frac 1{x^2}=$ $\frac Ax + B{x^2}$ ? Or was that a typo for $\frac {A}{x} + \frac{B}{x^2}$ ?

If it was the first one I failed my test miserably.

yes, it was a typo. i just caught it. i fixed it

**Jhevon** · May 4th 2008, 01:44 PM

Originally Posted by Krizalid

$\begin{aligned} \frac{1}{(x+2)\left( x^{2}+4 \right)}&=\frac{\left( x^{2}+4 \right)-\left( x^{2}-4 \right)}{8(x+2)\left( x^{2}+4 \right)} \\ & =\frac{1}{8}\left\{ \frac{1}{x+2}-\frac{x-2}{x^{2}+4} \right\} \\ & =\frac{1}{8}\left\{ \frac{1}{x+2}-\frac{x}{x^{2}+4}+\frac{2}{x^{2}+4} \right\}. \end{aligned}$

Integrate.

How do you see these things?!

i wanted to do algebraic manipulation to get it, but i didn't see that...

**Krizalid** · May 4th 2008, 01:48 PM

$x^2+4$ helped, since it's located in the denominator. Then I thought in $x^2-4,$ after subtraction I saw that I was doing worked out, besides $x^2-4$ it's the difference of perfect squares. Finally by seeing the $x+2$ in the denominator I could make it.

I'm sorry if this explanation looks terrible, but it's pretty hard to explain what I did in a couple of seconds.

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