Question

How do you calculate the antiderivative of `(-2x^3+14x^9)/(x^-2)`?

Answer 1

`-1/3x^6+7/6x^12+C`

Explanation:

Recall that `(a+b)/c=a/c+b/c`

Thus, we can rewrite `(-2x^3+14x^9)/x^-2` as `(-2x^3)/x^-2+(14x^9)/x^-2=-2x^(3-(-2))+14x^(9-(-2))=-2x^5+14x^11`

So, we want

`int(-2x^5+14x^11)dx`

Split this up, as we can split up sums or differences when integrating:

`int-2x^5dx+int14x^11dx`

Factor out the constants:

`-2intx^5dx+14intx^11dx`

Now, recall `intx^adx=x^(a+1)/(a+1)+C` where `C` is the constant of integration, just an arbitrary constant.

So, integrating, we get

`(-2x^6)/6+(14x^12)/12=-1/3x^6+7/6x^12+C`

Yes, we would technically have two constants of integration as we had two integrals, but we absorb them all into one constant.