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

Explanation:

Step1: Use substitution

Let $u = 1 + x$, then $du=dx$. When $x = 0$, $u = 1$; when $x = 8$, $u=9$.

Step2: Rewrite the integral

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

Step3: Apply power - rule for integration

The antiderivative of $u^{-\frac{1}{2}}$ is $2u^{\frac{1}{2}}+C$.

Step4: Evaluate the definite integral

$2u^{\frac{1}{2}}\big|_{1}^{9}=2\times9^{\frac{1}{2}}-2\times1^{\frac{1}{2}}$. $2\times3 - 2\times1=6 - 2$.

Answer:

$4$