Question

How do you integrate `(x^3+2x^2-x) / x dx`?

Answer 1

`x^3/3+x^2-x+C`

Explanation:

At first we simplify the integrand

`(x^3+2x^2-x)/x=x^3/x+2x^2/x-x/x=x^2+2x-1`
Now we can integrate the result:

`int(x^2+2x-1)dx=x^3/3+2x^2/2-x+C=x^3/3+x^2-x+C`

we have used that

`int x^ndx=x^(n+1)/(n+1)+C` if `n ne -1`