# `f(x) = root(3)(x^2) + xsqrt(x)` Find the most general antiderivative of the function. (Check your answer by differentiation.)

*print*Print*list*Cite

Expert Answers

lemjay | Certified Educator

`f(x)= root(3)(x^2) + xsqrtx`

To determine the most general antiderivative of this function, take the integral of f(x).

`F(x) = intf(x) dx = int (root(3)(x^2)+xsqrtx)dx`

Express the radicals in exponent form.

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

Then, apply the power formula of integral, `int u^ndu=u^(n+1)/(n+1) +C` .

`=x^(5/3)/(5/3)+x^(5/2)/(5/2)+C = (3x^(5/3))/5+(2x^(5/2))/5+C=(3x*x^(2/3))/5+(2x^2x^(1/2))/5+C`

`=(3xroot(3)(x^2))/5+(2x^2sqrtx)/5+C`

**Thus, the general antiderivative of `f(x)=root(3)(x^2)+xsqrtx` is `F(x)=(3xroot(3)(x^2))/5+(2x^2sqrtx)/5+C` .**