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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Help!

How do I integrate x^2/4-x?

Thank you!

Study Forum Helper

Original post by Hlrk

How do I integrate x^2/4-x?

Thank you!

Carry out some sort of division and integrate term by term.

OP

How would I go about that?

Study Forum Helper

Original post by Hlrk

How would I go about that?

Using your preferred method of polynomial division.

Study Forum Helper

If you can't remember how to divide polynomials you could note that:

\displaystyle \frac{x^2}{4-x}=\frac{x^2-16+16}{4-x}=\frac{(x+4)(x-4)+16}{4-x}=-(x+4)+\frac{16}{4-x}

Original post by Hlrk

How do I integrate x^2/4-x?

Thank you!

An alternative method would be to substitute u=4-x

Kvothe the Arcane

Original post by Mr M

If you can't remember how to divide polynomials you could note that:

\displaystyle \frac{x^2}{4-x}=\frac{x^2-16+16}{4-x}=\frac{(x+4)(x-4)+16}{4-x}=-(x+4)+\frac{16}{4-x}

Huh...I have never approached it like that!

Original post by keromedic

Huh...I have never approached it like that!

It's just another way of approaching the problem of rewriting the integrand in a simpler way - polynomial long division will always work, but if you can spot a trick like this it always helps!

Kvothe the Arcane

Original post by davros

It's just another way of approaching the problem of rewriting the integrand in a simpler way - polynomial long division will always work, but if you can spot a trick like this it always helps!

Thanks
I get how this works. I hate long division so kind of do what MrM did but rather unintuively Through use of the remainder theorem.
Think it goes something like

\dfrac{P(x)}{Q(x)}=R(x)+\dfrac{S(x)}{Q(x)}

..

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