Question

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

Explanation:

Step1: Use substitution

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

Step2: Apply power - rule for integration

The power - rule for integration is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). For $\int\frac{du}{\sqrt{u}}=\int u^{-\frac{1}{2}}du$. Using the power - rule, we get $\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2u^{\frac{1}{2}}+C$.

Step3: Evaluate the definite integral

$2u^{\frac{1}{2}}\big|_{1}^{9}=2\sqrt{9}-2\sqrt{1}$.

Step4: Calculate the result

$2\times3 - 2\times1=4$.

Answer:

$4$