Question

How do you evaluate the indefinite integral `int (-3x^2-x+6)dx`?

Answer 1

I got:`-x^3-x^2/2+6x+c`

Explanation:

First I would break it into smaller integrals:
`int(-3x^2)dx-intxdx+int6dx=`
and take the constant ot of the integral sign:
`=-3int(x^2)dx-intxdx+6intx^0dx=`
[where in the last one I used the fact that `x^0=1`]

then use the fundamental expression for the integral of a power:

`color(red)(intx^ndx=(x^(n+1))/(n+1)+c)`

and solve it as:

`=-cancel(3)x^3/cancel(3)-x^2/2+6x+c`