Question

$int_{0}^{8} \frac{dx}{sqrt{1+sqrt3{x}}}$

Explanation:

Step1: Simplify the integrand

$\frac{\sqrt{x}}{x} = x^{\frac{1}{2}} \cdot x^{-1} = x^{-\frac{1}{2}}$

Step2: Apply power rule for integration

The power rule for integration is $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n
eq -1$). Here $n=-\frac{1}{2}$, so:
$\int x^{-\frac{1}{2}} dx = \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = 2\sqrt{x} + C$

Step3: Evaluate definite integral from 1 to 8

$\int_{1}^{8} x^{-\frac{1}{2}} dx = 2\sqrt{x}\bigg|_{1}^{8} = 2\sqrt{8} - 2\sqrt{1}$

Step4: Simplify the result

$2\sqrt{8} = 2 \cdot 2\sqrt{2} = 4\sqrt{2}$, and $2\sqrt{1}=2$, so:
$4\sqrt{2} - 2$

Answer:

$4\sqrt{2} - 2$