how do you find f(x) of f'(x)=(x^4-x^2)/x^6
how about take the derivative?
i mean integral
Yeah how do I do that?
\[f'(x) = \frac{x^4 - x^2}{x^6}\] \[\int f'(x) = \int \frac{x^4 - x^2}{x^6}\] \[f(x) = \int \frac{x^4 - x^2}{x^6}\] \[f(x) = \int \frac{x^4}{x^6} - \frac{x^2}{x^6}\] getting an idea now?
oops minor mistake \[f(x) = \int \frac{x^4}{x^6} - \int \frac{x^2}{x^6}\] you might get confused
not really i'm kinda stuck after that part.
how about simplifying it first? \[\large \frac{x^m}{x^n} \implies x^{m-n}\] remember that rule?
\[ f(x)=\int x^{4-6}- \int x^{2-6}\\f(x)=\int x^{-2}- \int x^{-4} \]
haha and then after that part? sorry I just started this stuff and i'm really confused.
And these could be written as\[ f(x)=\int \frac{1}{x^2}- \int \frac{1}{x^4}\]
omg thank you i think i get it now.
Then simply use the fact that \[\large \int x^n dx= \frac{x^{(n+1)}}{(n+1)} + C\]
pellet how do I apply that?
Lol pellet xD
@lgbasallote Would explain it better than me :)
wtf. it said something different'
It's a filter
like @zepp said \[f(x) = \int x^{-2} - \int x^{-4}\] apply the rule on each one
And use substitution
\[\int x^{-2}\] your n here is -2 so use \[\int x^n = \frac{x^{n+1}}{n+1}\]
i have to go now so sorry i cannot explain further but here's an example to guide you \[\large \int x^{-3} \implies \frac{x^{-3 + 1}}{-3} \implies \frac{x^{-2}}{-2} \implies -\frac{1}{2x^2}\]
^note that's not the answer that was a sample
why did you change the 3rd step to -2 down the bottem?
uhmm sorry that was supposed to be \[\frac{x^{-3 + 1}}{-3 + 1}\] that's how it became -2 in the denominator
hope you got that correction \[\int x^{-3} = \frac{x^{-3+1}}{-3+1}\]
okay i need to go now. good luck!
oh okay thank you :)
\[\large \begin{align} f(x)&= \int x^{-2} - \int x^{-4} \\ f(x)&= \frac{x^{(-2+1)}}{-2}-\frac{x^{(-4+1)}}{-4}\\ f(x)&= \frac{x^{-1}}{-1} - \frac{x^{-3}}{-4}\\ f(x)&= \frac{\frac{1}{x}}{-1}-\frac{\frac{1}{x^3}}{-4}\\ f(x)&=-\frac{1}{x}-\frac{1}{-4x^3}\\ f(x)&=-\frac{1}{x}+\frac{1}{4x^3}+C\\ \end{align}\]
Got it? :D
haha yes I finally do. thank you very much guys :)
Good :)
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