Integral of [x+3/9-x²]dx?

1 Answer
Apr 21, 2018

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