Question

The value?

∫(x^4 - 1/x^2)dx

Answer 1

`x^5/5+1/x+C`

Explanation:

Since `int x^n = x^{n+1}/{n+1}+C`, this integral is
`int (x^4 - 1/x^2)dx = int x^4 dx-int x^-2 dx = x^{4+1}/{4+1}-x^{-2+1}/{-2+1} +C= x^5/5+1/x+C`

Answer 2

You are not looking for a "value" you are looking for a function called primitive. Se details below

Explanation:

`int (x^4-1/x^2)dx=`

By linearity of integral we have

`int (x^4-1/x^2)dx=intx^4dx - int 1/x^2dx`

We know that the primitive of `intx^ndx=1/(n+1)x^(n+1)+C` where `C` is a arbitrary constant and `n !=1`. For that reason, the result is

`intx^4dx - int 1/x^2dx=1/5x^5+1/x+C`

We can proof that this is the searched function appliying derivatives. Lets see:

`f(x)=1/5x^5+1/x+C`

`f'(x)=1/5·5x^(5-1)+(-1/x^2)+0=x^4-1/x^2`