Questions: Evaluate the integral. [ int frac1sqrt3 s+1 d s int frac1sqrt3 s+1 d s= ]

$Evaluate the integral. [ int frac1sqrt3 s+1 d s int frac1sqrt3 s+1 d s= ]$

Transcript text: Evaluate the integral. \[ \begin{array}{l} \int \frac{1}{\sqrt{3 s+1}} d s \\ \int \frac{1}{\sqrt{3 s+1}} d s= \end{array} \]

Solution

Solution Steps

To evaluate the integral $\int \frac{1}{\sqrt{3s+1}} \, ds$, we can use a substitution method. Let $u = 3s + 1$. Then, $du = 3 \, ds$ or $ds = \frac{1}{3} \, du$. This substitution will simplify the integral into a more manageable form.

Step 1: Substitution

To evaluate the integral $\int \frac{1}{\sqrt{3s+1}} \, ds$, we start by making a substitution. Let $u = 3s + 1$. Then, $du = 3 \, ds$ or $ds = \frac{1}{3} \, du$.

Step 2: Simplify the Integral

Substitute $u$ and $ds$ into the integral: \[ \int \frac{1}{\sqrt{3s+1}} \, ds = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{3} \, du = \frac{1}{3} \int u^{-\frac{1}{2}} \, du \]

Step 3: Integrate

Now, integrate $u^{-\frac{1}{2}}$: \[ \frac{1}{3} \int u^{-\frac{1}{2}} \, du = \frac{1}{3} \cdot 2u^{\frac{1}{2}} = \frac{2}{3} \sqrt{u} \]

Step 4: Substitute Back

Replace $u$ with $3s + 1$: \[ \frac{2}{3} \sqrt{u} = \frac{2}{3} \sqrt{3s + 1} \]