Question

Another partial fractions question

Answer 1

One second while I type out what I have

Answer 2

\[\int\limits_{}^{}\frac{ 3x^4-2x^3+2x^2+5 }{ x^2(x-1) }\]

is the question.

Answer 3

I assume you know you have to first divide

Answer 4

\[3x^4-2x^3+2x^2+5=\frac{ Ax+B }{ x^2 }+\frac{ C }{ (x-1) } \rightarrow (Ax+B)(x-1)+Cx^2\]

A=-3
B=-5
C=8

Answer 5

Oh I see @phi

Answer 6

you have to divide first. you need the top to have lower order than the bottom

Answer 7

Could I have continued doing it the way I was?

Answer 8

I don't think so.

Answer 9

Oh, is that just something to remember? If the top is higher order you cant integrate as long as there are variables in the denominator?

Answer 10

you divide to get a quotient (which you can integrate)
and a remainder/ divisor which you can use partial fractions on

Answer 11

how can I tell how many times \[(x^3-x^2)\div3x^4\] using long division?

Answer 12

Sorry basic algebra question, it's been a while since I've had to use long division with polynomials

Answer 13

yes, I would use long division

Answer 14

\(\dfrac{x^3-x^2}{3x^4}\) is already as simple as it gets as far as long division goes. You can still reduce it to \(\dfrac{1}{3x}-\dfrac{1}{3x^2}\) though.

Answer 15

I'll leave it to OP to decide if this alternative method seems easier.

One thing you can do is rearrange the integrand like so,
\[\begin{align*}
\frac{3x^4-2x^3+2x^2+5}{x^2(x-1)}&=\frac{3x^4-2x^3+2x^2}{x^2(x-1)}+\frac{5}{x^2(x-1)}\\[1ex]
&=\frac{3x^2-2x+2}{x-1}+\frac{5}{x^2(x-1)}
\end{align*}\]The first can have its numerator rewritten with a shift in \(x\):
\[\begin{align*}
3x^2-2x+2&=a(x-1)^2+b(x-1)+c\\[1ex]
&=ax^2-2ax+1+bx-b+c\\[1ex]
&=ax^2+(b-2a)x+(1-b+c)
\end{align*}\]Matching up your terms' coefficients, you end up having to solve the system
\[\begin{cases}a=3\\b-2a=-2\\1-b+c=2\end{cases}\implies \begin{cases}a=3\\b=4\\c=3\end{cases}\]and so now you can simplify the first term:
\[\frac{3x^2-2x+2}{x-1}=\frac{3(x-1)^2+4(x-1)+3}{x-1}=3(x-1)+4+\frac{3}{x-1}\]
Putting everything together, we've basically rewritten the integral as
\[\int\frac{3x^4-2x^3+2x^2+5}{x^2(x-1)}\,\mathrm{d}x=\int\left(3(x-1)+4+\frac{3}{x-1}+\frac{5}{x^2(x-1)}\right)\,\mathrm{d}x\]where you can proceed with the last term via PFD.