# Thread: Integration by Partial Fractions

1. ## Integration by Partial Fractions

$
\int {\frac{{2x^4 }}
{{x^3 - x^2 + x - 1}}dx}
$

This one gets a little bit tricky. So first off I do long division to make the fraction proper:

$
x^3 - x^2 + x - 1\left){\vphantom{1{2x^4 }}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{2x^4 }}}
$

Here is my result. You can check my division although i'm not skilled enough in MathType to show all of it yet.

$
\int {\left( {2x + 2 + \frac{2}
{{x^3 - x^2 + x - 1}}} \right)} dx
$

Now I need to find out a way to simplify that last term.

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

This is as far as I went. Any help is appreciated. Thanks!

2. Hello,

As for the last one... Look at the polynomial $x^3-x^2+x-1$. You can see that the sum of the coefficients is 0. So 1 is a root of it.
Hence you can factor by $(x-1)$
Actually, after some calculations, $x^3-x^2+x-1=(x-1)(x^2+1)$

And you can do partial fractions here

3. Originally Posted by Moo
Hello,

As for the last one... Look at the polynomial $x^3-x^2+x-1$. You can see that the sum of the coefficients is 0. So 1 is a root of it.
Hence you can factor by $(x-1)$
Actually, after some calculations, $x^3-x^2+x-1=(x-1)(x^2+1)$

And you can do partial fractions here
Great. The rest is simple. I was trying to come up with some similar but had no luck. I don't quite understand what you mean one being the root of it. Could you further explain this to me?

thanks Moo once again.

4. Originally Posted by RedBarchetta
Great. The rest is simple. I was trying to come up with some similar but had no luck. I don't quite understand what you mean one being the root of it. Could you further explain this to me?

thanks Moo once again.
A root is a number that annulates a polynomial.

Let $P(x)=x^3-x^2+x-1$

If the sum of the coefficients is 0, you can say that 1 is a root, that is to say $P(1)=0$ : $P(1)=1^3-1^2+1-1=0$

More generally, let a polynomial $Q(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$, with the sum of coefficients equal to 0.

If we let x=1, we'll have $Q(1)=a_n+a_{n-1}+\dots+a_1+a_0$ and this equals 0 since the sum of all coefficients is 0. $Q(1)=0$ hence 1 is a root.

Now, if a is a root of a polynomial Q(x), then Q(x) is a multiple of (x-a).

It's a bit messy, but it should be all I can say on the subject :s

5. Originally Posted by RedBarchetta
$
\int {\frac{{2x^4 }}
{{x^3 - x^2 + x - 1}}dx}
$
First $x^3-x^2+x-1=x^2(x-1)+(x-1)=(x-1)(x^2+1).$ Now (I omitted the constant 2, nothing out the extraordinary),

\begin{aligned}\int\frac{x^4}{x^3-x^2+x-1}&=\int\frac{(x^2+1)(x^2-1)+1}{(x-1)(x^2+1)}\,dx\\&=\int(x+1)\,dx+\frac12\int\frac{x ^2+1-(x^2-1)}{(x-1)(x^2+1)}\,dx\\&=\int(x+1)\,dx+\frac12\left(\int\ frac{dx}{x-1}\,dx-\int\frac{x+1}{x^2+1}\,dx\right).\end{aligned}

Finish off from there, gtg now.