Finding function when dx is integrated $\int (5-x^2+\frac{18}{x^4})dx$ I have a question on how to evaluate this integral:
$\int (5-x^2+\frac{18}{x^4})dx$
Is this correct?
$\int (5-x^2+\frac{18}{x^4})dx$
$ \int 5 dx = 5x + C$ 
$ \int -x^2 = \frac {-x^3}{3}= -\frac{1}{3}x^3 +C $
$ \int \frac {18}{x^4} = \int 18x^{-4} dx = \frac {18x^{-3}}{3}=6x^{-3} + C$ 

So, $\int (5-x^2+\frac{18}{x^4})dx= 5x-\frac{1}{3}x^3+6x^{-3}+C$

It is not quite correct.  Note that 
$$\int 18x^{-4}\,dx=18\,\left(\frac{x^{(-4+1)}}{(-4+1)}\right)+C=-6x^{-3}+C$$

On a side note, it is not good form to write "$5=5x$" as short hand notation for "$\int 5\,dx=5x$."

