Question

Integrate

4/x^3 dx

Answer 1

\[\int\limits_{}^{}\frac{ 4 }{ x^3 } dx\]

Answer 2

\[\large\rm \int\limits \frac{4}{x^3}dx\quad=\int\limits 4x^{-3}dx\]Just power rule from here, ya? :)

Answer 3

Whats the power rule again?

Answer 4

\[\large\rm \int\limits x^{n}dx\quad= \frac{1}{n+1}x^{n+1}\]

Answer 5

plus C (there's a constant of integration)

Answer 6

When you apply power rule for differentiation,
you apply these steps in this order:
~power comes down to multiply
~power on x decreases by 1.

So when you apply power rule for integration,
it's the opposite steps,
applied in the opposite order:
~increase power on x by 1,
~new power comes down to divide

Answer 7

i did the power rule and got \[\frac{ 1 }{ -2 }x^-2\]

Answer 8

Hmm your power looks correct.\[\large\rm \int\limits 4x^{-3}dx\quad=\frac{4}{-3+1}x^{-3+1}\]We still have the 4 though, ya? :)

Answer 9

Wait why isn't the numerator 1?

Answer 10

\[\large\rm \color{orangered}{4}\cdot \int\limits\limits x^{-3}dx\quad=\color{orangered}{4}\cdot\frac{1}{-3+1}x^{-3+1}\]Maybe we would bring the 4 outside of the integral as a first step.
Do you see why we still have the 4?

Answer 11

yes

Answer 12

The power rule only showed us how to deal with this:\[\large\rm \int\limits x^{-3}dx\quad=-\frac{1}{2}x^{-2}\]But we had an extra 4 in the mix, ya? :)\[\large\rm \int\limits\limits 4x^{-3}dx\quad=-\frac{4}{2}x^{-2}\]

Answer 13

yes

Answer 14

And then I guess you can divide the 4 and 2,
and maybe put the x back in the denominator at this point.

Answer 15

And ya, don't forget the constant of integration :)

Answer 16

What do you think?
Confused?
Brain still spinnin? :d

Answer 17

i get it