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

Explanation:

Step1: Substitute $u=1+x$

Let $u = 1+x$, then $du = dx$. When $x=0$, $u=1$; when $x=8$, $u=9$. The integral becomes $\int_{1}^{9} \frac{du}{\sqrt{u}} = \int_{1}^{9} u^{-\frac{1}{2}} du$

Step2: Apply power rule for integration

The power rule is $\int u^n du = \frac{u^{n+1}}{n+1} + C$ for $n\neq-1$. Here $n=-\frac{1}{2}$, so: $\int_{1}^{9} u^{-\frac{1}{2}} du = \left. 2u^{\frac{1}{2}} \right|_{1}^{9}$

Step3: Evaluate definite integral

Calculate at bounds: $2(9^{\frac{1}{2}}) - 2(1^{\frac{1}{2}}) = 2\times3 - 2\times1$

Answer:

$4$