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

Question

lgbasallote · Answer

how about take the derivative?

lgbasallote · Answer

i mean integral

anonymous · Answer

Yeah how do I do that?

lgbasallote · Answer

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

lgbasallote · Answer

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

anonymous · Answer

not really i'm kinda stuck after that part.

lgbasallote · Answer

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

zepp · Answer

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

anonymous · Answer

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

zepp · Answer

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

anonymous · Answer

omg thank you i think i get it now.

zepp · Answer

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

anonymous · Answer

pellet how do I apply that?

zepp · Answer

Lol pellet xD

zepp · Answer

@lgbasallote Would explain it better than me :)

anonymous · Answer

wtf. it said something different'

zepp · Answer

It's a filter

lgbasallote · Answer

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

zepp · Answer

And use substitution

lgbasallote · Answer

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

lgbasallote · Answer

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

lgbasallote · Answer

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

anonymous · Answer

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

lgbasallote · Answer

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

lgbasallote · Answer

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

lgbasallote · Answer

okay i need to go now. good luck!

anonymous · Answer

oh okay thank you :)

zepp · Answer

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

zepp · Answer

Got it? :D

anonymous · Answer

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

zepp · Answer

Good :)