# Thread: Integration of a Rational Function

1. ## Integration of a Rational Function

I have a problem that i don't understand what the solutions manual did. I am learning as much calculus 2 over the summer as i can.

$\displaystyle \int \frac {x^3}{x^2 + 1}\,dx$

Here is the part where i am confused..

$\displaystyle \int [ {x} - \frac {x}{x^2 + 1}]\,dx$

Then,

$\displaystyle \int {x} \,dx - \frac {1}{2} \int \frac {2x}{x^2 +1} \,dx$

Finally
$\displaystyle \frac {1}{2} \, {x^2} - \frac {1}{2} ln(x^2 + 1) + c$
(using long division)

I am not sure how they got the answer by the steps. The first step is the most confusing. the second step i can see how they got it. Third step is confusing on how they got

$\displaystyle \frac {1}{2} \, {x^2}$.
The rest of it i understand.
Can anyone explain the first step and the third one?

2. ## re: Integration of a Rational Function

x^3/(x^2+1)=(x^3+x-x)/(x^2+1)=(x^3+x)/(x^2+1)-x/(x^2+1)=x-x/(x^2+1)

3. ## re: Integration of a Rational Function

First step is just a long division

or it could be done as follows :

$\displaystyle \displaystyle \frac{x^3}{x^2+1} = x \cdot \frac{x^2}{x^2+1} = x \cdot \frac{x^2+1-1}{x^2+1} = x \cdot \left( \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} \right)$

$\displaystyle \displaystyle = x \cdot \left( 1 - \frac{1}{x^2+1} \right) = x - \frac{x}{x^2+1}$

and $\displaystyle \frac{1}{2} \, x^2$ is just the integral of x.

4. ## re: Integration of a Rational Function

In general:
$\displaystyle \int x^ndx =\frac{x^{n+1}}{n+1}+C$ if $\displaystyle n\neq -1$
Use this formule to calculate $\displaystyle \int x^2dx$

5. ## re: Integration of a Rational Function

$\displaystyle \int\ x\ dx=\frac{1}{2}x^2$

because

$\displaystyle \frac{d}{dx}\frac{1}{2}x^2=\frac{1}{2}\cdot2x=1 \cdot x=x$

6. ## re: Integration of a Rational Function

Ok, thanks for the reply. Why do you use a negative one here x^2 +1 - 1 ?

7. ## re: Integration of a Rational Function

Thanks everyone.

$\displaystyle \frac {1}{2} \, {x^2}$.

is just the anti dirivitive. i can see it now!!!!

8. ## re: Integration of a Rational Function

Originally Posted by icelated
Ok, thanks for the reply. Why do you use a negative one here x^2 +1 - 1 ?
If you choose General's way, then you've to subtract -1 if you add +1 to not make a mistake. I mean:
$\displaystyle \frac{x^3}{x^2+1}=x\cdot \frac{x^2}{x^2+1}=x\cdot \frac{x^2+1-1}{x^2+1}$
Because the original numerator is $\displaystyle x^2$ so if you add +1 then you have to subtract -1 to hold the original numerator $\displaystyle x^2$.

Thanks