<s>[INST] 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, $\int (5-x^2+\frac{18}{x^4})dx= 5x-\frac{1}{3}x^3+6x^{-3}+C$
 [/INST] This is how you would do this
$$\int (5-x^2+\frac{18}{x^4})dx$$ 
Integration as you have probably shown spits across addition so we have;
$$
\int5dx+\int-x^2dx+\int\frac{18}{x^4}dx
$$
now we can also take out constaints so we have:
$$
5\int1dx-\int x^2dx+18\int x^{-4}dx
$$
now by defintion we have
$$
5x-\frac{x^3}{3}+18\frac{x^{-3}}{-3}+C
$$
Not we only need one C because three constants added together are constants
$$
5x-\frac{x^3}{3}-\frac{6}{x^3}+C
$$
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