Thread: quick integral conversion question

1. quick integral conversion question

Im looking at the solution to one of my teachers answers and just wondering how he got this:

(X^4) / ({X^2}+1)dx is the same statement as X^2-1+[1/({X^2}+1)]dx

Both are integrals obviously and i dont know how he got the 2nd from the 1st. Some sort of algebra/identity i assume? Thanks

Chris

2. Originally Posted by chrsr345
Im looking at the solution to one of my teachers answers and just wondering how he got this:

(X^4) / ({X^2}+1)dx is the same statement as X^2-1+[1/({X^2}+1)]dx

Both are integrals obviously and i dont know how he got the 2nd from the 1st. Some sort of algebra/identity i assume? Thanks

Chris
Put the three terms in second expression under a common denominator and you'll get the first.

The process of going form the first to the second is a bit more difficult, but it's called breaking a fraction into partial fractions.
You basically do it by establishing an equation like:

$\displaystyle \frac{x^4}{x^2+1} = A + \frac{B}{x^2+1}$,

then multiply by the denominator to get:

$\displaystyle x^4 = A(x^2+1) + B$

Expand:

$\displaystyle x^4 = Ax^2+(A + B)$

and compare terms on each side.
Get $\displaystyle A = x^2$ and $\displaystyle A+B=0$, so $\displaystyle B = -x^2$.
Therefore
$\displaystyle \frac{x^4}{x^2+1} = x^2- \frac{x^2}{x^2+1}$

This process can be repeated on the remaining fraction:

$\displaystyle \frac{x^2}{x^2+1} = C + \frac{D}{x^2 + 1}$.
Multiply to get:

$\displaystyle x^2 = C(x^2+1)+D = Cx^2 + (C+D)$.

Compare terms again, and this time get $\displaystyle C = 1$, and $\displaystyle D = -1$.

Put it all together:

$\displaystyle \frac{x^4}{x^2+1} = x^2- 1 + \frac{1}{x^2+1}$

3. You don't even need partial fractions, you just need add & remove:

$\displaystyle \frac{{x^4 }} {{x^2 + 1}} = \frac{{\left( {x^4 - 1} \right) + 1}} {{x^2 + 1}} = \frac{{\left( {x^2 + 1} \right)\left( {x^2 - 1} \right) + 1}} {{x^2 + 1}} = x^2 - 1 + \frac{1} {{x^2 + 1}}.$

4. Originally Posted by Krizalid
You don't even need partial fractions, you just need add & remove:

$\displaystyle \frac{{x^4 }} {{x^2 + 1}} = \frac{{\left( {x^4 - 1} \right) + 1}} {{x^2 + 1}} = \frac{{\left( {x^2 + 1} \right)\left( {x^2 - 1} \right) + 1}} {{x^2 + 1}} = x^2 - 1 + \frac{1} {{x^2 + 1}}.$
Much slicker!

5. I prefer this rather than partial fractions, it takes less time. (Of course, not always it's possible to avoid such technique.)

6. Thanks alot guys! Kind of rediculous my teacher did not elaborate on this at all. Cheers! Kanpai! Konbae!