Integration by partial fractions

[klooless]

Hello, I'm working on this problem;

?(x^4 - 2x^2 + 4x +1) / (x^3 - x^2 - x + 1) dx

I figured out the long division to be x+1+ (4x)/(x^3 - x^2 - x + 1).

I'm stumped how to factor the denominator... help please?

cheers!

 

[galactus]

Rewrite \(\displaystyle x^{3}-x^{2}-x+1=x^{3}+x^{2}-2x^{2}-2x+x+1\)

Group: \(\displaystyle (x^{3}-2x^{2}+x)+(x^{2}-2x+1)\)

Factor: \(\displaystyle x(x^{2}-2x+1)+(x^{2}-2x+1)\)

\(\displaystyle (x+1)(x-1)^{2}\)


You should get \(\displaystyle \frac{1}{x-1}-\frac{1}{x+1}+\frac{2}{(x-1)^{2}}+x+1\)

Now integrate.

 

[Deleted member 4993]

Hello, I'm working on this problem;

?(x^4 - 2x^2 + 4x +1) / (x^3 - x^2 - x + 1) dx

I figured out the long division to be x+1+ (4x)/(x^3 - x^2 - x + 1).

I'm stumped how to factor the denominator... help please?

cheers!

By observation (or by using rational root theorem) you can see f(1) = 0 (or x=1 is a root of the polynomial.

So (x-1) is a factor.

Now do long division by (x-1) to get:

x[sup:36wkyen0]3[/sup:36wkyen0] - x[sup:36wkyen0]2[/sup:36wkyen0] - x + 1 = (x - 1)(x[sup:36wkyen0]2[/sup:36wkyen0] - 1)

finally,

x[sup:36wkyen0]3[/sup:36wkyen0] - x[sup:36wkyen0]2[/sup:36wkyen0] - x + 1 = (x - 1)(x[sup:36wkyen0]2[/sup:36wkyen0] - 1) = (x - 1)(x + 1)(x - 1) = (x - 1)[sup:36wkyen0]2[/sup:36wkyen0](x + 1)

Now do partial fraction part....

 

[BigGlenntheHeavy]

\(\displaystyle \int\frac{x^{4}-2x^{2}+4x+1}{x^{3}-x^{2}-x+1}dx \ = \ \int(\frac{-1}{x+1}+\frac{1}{x-1}+\frac{2}{(x-1)^{2}}+x+1)dx\)

Now integrate, dividing through isn't necessary, also the partial fractions can be gotten by expand on your TI-89 (F2-3).

 

[klooless]

Thanks Glenn,

Cheers!