evaluate the integral radical x^3, i got answer 2(x^3)^3/2 over 3
evaluate integral x^3-x^2/x^2? how do i do it
Your answer to the first is incorrect. Notice that $\displaystyle \displaystyle \sqrt{x^3} = x^{\frac{3}{2}}$. Apply the power rule.
For the second, simplify $\displaystyle \displaystyle \frac{x^3 - x^2}{x^2} = \frac{x^2 (x - 1)}{x^2} = x - 1$. Surely you can integrate that...