Question

Integral of [x+3/9-x²]dx?

Answer 1

`1/2x^2+3/9x-1/3x^3+C`

Explanation:

To take the integral of this expression, you should use the reverse power rule.

`inta^n=(a^(n+1))/(n+1)`

In this case, you have three terms: `x`, `3/9`, and `-x^2`

The integral of `x` by itself is

`intx^1=(x^(1+1))/(1+1)=x^2/2`

`3/9` which should be seen as `3/9*x^0` has the integral

`int3/9*x^0=3/9x^(0+1)/(0+1)=3/9x`

Finally, the integral of `-x^2` is

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

Combining all these terms and an unknown constant (C) you get the final answer:

`int(x+3/9-x^2)dx=1/2x^2+3/9x-1/3x^3+C`