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Mathematics 68 Online
OpenStudy (anonymous):

how do you find f(x) of f'(x)=(x^4-x^2)/x^6

OpenStudy (lgbasallote):

how about take the derivative?

OpenStudy (lgbasallote):

i mean integral

OpenStudy (anonymous):

Yeah how do I do that?

OpenStudy (lgbasallote):

\[f'(x) = \frac{x^4 - x^2}{x^6}\] \[\int f'(x) = \int \frac{x^4 - x^2}{x^6}\] \[f(x) = \int \frac{x^4 - x^2}{x^6}\] \[f(x) = \int \frac{x^4}{x^6} - \frac{x^2}{x^6}\] getting an idea now?

OpenStudy (lgbasallote):

oops minor mistake \[f(x) = \int \frac{x^4}{x^6} - \int \frac{x^2}{x^6}\] you might get confused

OpenStudy (anonymous):

not really i'm kinda stuck after that part.

OpenStudy (lgbasallote):

how about simplifying it first? \[\large \frac{x^m}{x^n} \implies x^{m-n}\] remember that rule?

OpenStudy (zepp):

\[ f(x)=\int x^{4-6}- \int x^{2-6}\\f(x)=\int x^{-2}- \int x^{-4} \]

OpenStudy (anonymous):

haha and then after that part? sorry I just started this stuff and i'm really confused.

OpenStudy (zepp):

And these could be written as\[ f(x)=\int \frac{1}{x^2}- \int \frac{1}{x^4}\]

OpenStudy (anonymous):

omg thank you i think i get it now.

OpenStudy (zepp):

Then simply use the fact that \[\large \int x^n dx= \frac{x^{(n+1)}}{(n+1)} + C\]

OpenStudy (anonymous):

pellet how do I apply that?

OpenStudy (zepp):

Lol pellet xD

OpenStudy (zepp):

@lgbasallote Would explain it better than me :)

OpenStudy (anonymous):

wtf. it said something different'

OpenStudy (zepp):

It's a filter

OpenStudy (lgbasallote):

like @zepp said \[f(x) = \int x^{-2} - \int x^{-4}\] apply the rule on each one

OpenStudy (zepp):

And use substitution

OpenStudy (lgbasallote):

\[\int x^{-2}\] your n here is -2 so use \[\int x^n = \frac{x^{n+1}}{n+1}\]

OpenStudy (lgbasallote):

i have to go now so sorry i cannot explain further but here's an example to guide you \[\large \int x^{-3} \implies \frac{x^{-3 + 1}}{-3} \implies \frac{x^{-2}}{-2} \implies -\frac{1}{2x^2}\]

OpenStudy (lgbasallote):

^note that's not the answer that was a sample

OpenStudy (anonymous):

why did you change the 3rd step to -2 down the bottem?

OpenStudy (lgbasallote):

uhmm sorry that was supposed to be \[\frac{x^{-3 + 1}}{-3 + 1}\] that's how it became -2 in the denominator

OpenStudy (lgbasallote):

hope you got that correction \[\int x^{-3} = \frac{x^{-3+1}}{-3+1}\]

OpenStudy (lgbasallote):

okay i need to go now. good luck!

OpenStudy (anonymous):

oh okay thank you :)

OpenStudy (zepp):

\[\large \begin{align} f(x)&= \int x^{-2} - \int x^{-4} \\ f(x)&= \frac{x^{(-2+1)}}{-2}-\frac{x^{(-4+1)}}{-4}\\ f(x)&= \frac{x^{-1}}{-1} - \frac{x^{-3}}{-4}\\ f(x)&= \frac{\frac{1}{x}}{-1}-\frac{\frac{1}{x^3}}{-4}\\ f(x)&=-\frac{1}{x}-\frac{1}{-4x^3}\\ f(x)&=-\frac{1}{x}+\frac{1}{4x^3}+C\\ \end{align}\]

OpenStudy (zepp):

Got it? :D

OpenStudy (anonymous):

haha yes I finally do. thank you very much guys :)

OpenStudy (zepp):

Good :)

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