Question

How do you evaluate the indefinite integral `int (-4x^4+x^2-6)dx`?

Answer 1

The answer is `=-4/5x^5+1/3x^3-6x+C`

Explanation:

We use

`intx^ndx=x^(n+1)/(n+1)+C(x!=-1)`

Therefore,

`int(-4x^4+x^2-6)dx`

`=-4intx^4dx+intx^2dx-6intdx`

`=-4/5x^5+1/3x^3-6x+C`