Integral #4

Let $u = \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt[3]\cdots}}}$ . Then $u^3 = xu \implies u^2 = x \implies 2u\ du = dx$ and

$\int_1^3 \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt[3]\cdots}}} \ dx = \int_1^{\sqrt 3} 2u^2 \ du = \frac 23 u^3 \bigg|_1^{\sqrt 3} = \frac 23 (3\sqrt 3 - 1) = \sqrt{12} - \frac 23$

Therefore $a+b+c = 12 + 2 - \dfrac 23 = \dfrac {40}3 \implies p + q = 40 + 3 = \boxed{43}$ .

The integrand is $x^{\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...}=x^{\dfrac{1}{2}}$ . The value of the integral is thus $\dfrac{2}{3}(3^{\dfrac{3}{2}}-1)=\sqrt {12}-\dfrac{2}{3}$ . So $a=2,b=12,c=-\dfrac{2}{3}$ and $a+b+c=\dfrac{40}{3} \implies p=40, q=3$ and $p+q=\boxed {43}$ .