Integral 22

$\Large y=x^{\overbrace{\small{\frac{1}{2}-\frac{1}{4}+\frac{1}{8}+\cdots}}^{\color{#D61F06}{\text{Infinite GP}}}}$

$\Large =x^{\small{\frac{\frac 12}{1+\frac 12}}}=x^{\frac 13}$

Hence we have to find $\int_{0}^1 x^{\frac 13}\,dx$ . $\Large{ \int_0^1x^{\frac 13}\,dx}$ $\Large{=\dfrac{3}{4} x^{\frac 43}|_{0}^1=\dfrac 34 }$ $\huge \therefore ~3+4=\boxed 7$

$t=\sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\ddots}}}}}} \implies x=t^{3} \implies dx=3t^{2}dt$ $\int_{0}^{1}{3t^3}dt = \left [\frac{3t^4}{4} \right ]_{0}^{1} = \frac{3}{4} \\ 3+4=\boxed{7}$