Finding function when dx is integrated $\int (5-x^2+\frac{18}{x^4})dx$

Question

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

Answer 1

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 "$\int 5\,dx=5x$."

Answer 2

Well we know what you mean but the equal sign is obviously wrong. You should use the integral of sum rule and use an integral sign before each of the summands. Also I think the question was regarding $dx$. That's just a notation to specify the variable you're integrating with respect to.

Answer 3

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 $$ 5x-\frac{x^3}{3}-\frac{6}{x^3}+C $$