1. ## Partial Fraction Quadratic Factor Integration

I think I understand the basic idea behind partial fractions but I am stuck on a couple of problems with quadratic factors. I think if I saw one done I could figure out the rest. This is one of the ones giving me trouble. When I am done with the partial fraction I still seem unable to integrate it. I'm about a month behind in my online class and already have dropped it once so am hoping not to have to again.

2. Originally Posted by CrazyLond
I think I understand the basic idea behind partial fractions but I am stuck on a couple of problems with quadratic factors. I think if I saw one done I could figure out the rest. This is one of the ones giving me trouble. When I am done with the partial fraction I still seem unable to integrate it. I'm about a month behind in my online class and already have dropped it once so am hoping not to have to again.
Setting up our partial fractions we get

$\frac{x^2-x+9}{(x^2+9)^2}=\frac{Ax+B}{x^2+9}+\frac{Cx+D}{(x^ 2+9)^2}$

Multiplying through by $(x^2+9)^2$

we get

$x^2-x+9=(Ax+B)(x^2+9)+Cx+D=Ax^3+Bx^2+9Ax+Cx+9B+D$

Grouping we get $x^2-x+9=Ax^3+Bx^2+(9A+c)x+9B+D$

Equating Coefficients we first get

$Ax^3=0x^3\Rightarrow{A=0}$

Next we have that $Bx^2=x^2\Rightarrow{B=1}$

Next we have $9A+c=9\cdot{0}+C=C=-1$

And finally $9B+D=9+D=9\Rightarrow{D=0}$

Putting it back into our original we get

$\frac{x^2-x+9}{(x^2+9)^2}=\frac{1}{x^2+9}+\frac{-x}{(x^2+9)^2}$

So now $\int\frac{x^2-x+9}{(x^2+9)^2}dx=\int\bigg[\frac{1}{x^2+9}-\frac{x}{(x^2+
9)^2}\bigg]dx$

The first integral is a arctangent so

$\int\frac{dx}{(3)^2+x^2}=\frac{1}{3}arctan\bigg(\f rac{x}{3}\bigg)+C$

Noticing in the second one that x is 1/2 the derivative of the quantity

we rewrite to get

$\int\frac{x}{(x^2+9)^2}=\frac{1}{2}\int\frac{2x}{( x^2+9)^2}dx=\frac{-1}{2(x^2+9)}+C$

Combining we get

$\int\frac{x^2-x+9}{(x^2+9)^2}dx=\frac{1}{3}arctan\bigg(\frac{x}{ 3}\bigg)+\frac{1}{2(x^2+9)}+C$

3. Originally Posted by CrazyLond
I think I understand the basic idea behind partial fractions but I am stuck on a couple of problems with quadratic factors. I think if I saw one done I could figure out the rest. This is one of the ones giving me trouble. When I am done with the partial fraction I still seem unable to integrate it. I'm about a month behind in my online class and already have dropped it once so am hoping not to have to again.

No partial fractions are needed rewrite as
$\frac{x^2+9-x}{(x^2+9)^2}=\frac{x^2+9}{(x^2+9)^2}+\frac{-x}{(x^2+9)^2}=\frac{1}{x^2+9}-\frac{x}{(x^2+9)^2}$

4. Originally Posted by TheEmptySet
No partial fractions are needed rewrite as
$\frac{x^2+9-x}{(x^2+9)^2}=\frac{x^2+9}{(x^2+9)^2}+\frac{-x}{(x^+9)^2}$
Yeah but wasn't he asking for partial fractions decomposition?

5. Originally Posted by Mathstud28
Yeah but wasn't he asking for partial fractions decomposition?
Yes, but I think it is good for poeple to see other methods.

I didn't start "seeing" these for quite a long time, and I still miss them now.

This is just my opinion.

What do you think?

6. Originally Posted by TheEmptySet
Yes, but I think it is good for poeple to see other methods.

I didn't start "seeing" these for quite a long time, and I still miss them now.

This is just my opinion.

What do you think?

Well...I agree with you whole-heartdly...I was even about to do it that way...but I get chastised on here all the time for not answering what the poster directly asked....so I didnt

All in all I would say both ways are worth seeing

7. Thanks for the help. My problem was multiplying A and B by the square of what I should have.

The problem did ask for me to use partial fractions but thanks for showing that other method.