(int_{0}^{8} \frac{dx}{sqrt{1 + x}})

Answer:

$2(\sqrt{17}-3)$

Explanation:

Step1: Choose substitution

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

Step2: Rewrite integral

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

Step3: Apply power rule

Integrate: $\int u^n du = \frac{u^{n+1}}{n+1} + C$ for $n\neq-1$ $$\int_{1}^{9}u^{-\frac{1}{2}}du = \left.2u^{\frac{1}{2}}\right|_{1}^{9}$$

Step4: Evaluate definite integral

$$2\sqrt{9} - 2\sqrt{1} = 2\times3 - 2\times1 = 6 - 2 = 4$$ Correction note: Earlier substitution error fixed: When $x=0$, $u=0+1=1$; $x=8$, $u=8+1=9$. Re-evaluating correctly:

Step1: Correct substitution setup

Let $u = x + 1$, $du=dx$. Bounds: $x=0\to u=1$, $x=8\to u=9$.

Step2: Rewrite integral

$$\int_{1}^{9}u^{-1/2}du$$

Step3: Integrate using power rule

$$\left[2u^{1/2}\right]_{1}^{9}$$

Step4: Compute bounds

$$2\sqrt{9} - 2\sqrt{1} = 2(3) - 2(1) = 6 - 2 = 4$$