Question

Use the indicated formula from the table of integrals in this section to find the indefinite integral. $$ \int \frac{x}{\sqrt{2+3 x}} d x, \text { Formula } 19 $$

Short answer

Therefore, the indefinite integral \( \int \frac{x}{\sqrt{2+3x}} dx \) is \( \frac{2}{27}(2 + 3x)^\frac{3}{2} - \frac{4}{9}(2 + 3x)^\frac{1}{2} + C \)

Step-by-step solution

Identify the Problem

The problem lies in integrating the given function: \( \int \frac{x}{\sqrt{2+3x}} dx \). We need to identify a suitable substitution that will simplify the integral and make it match with Formula 19.

Apply Substitution

For the integral, let's substitute \( u = 2 + 3x \). Hence, the differential \( du = 3 dx \) and \( dx = \frac{du}{3} \). Also notice x can be expressed as \( x = \frac{u-2}{3} \) from \( u = 2 + 3x \). Substitute these values back to the integral.

Simplify the Integral

Replacing these values in given integral, we obtain \( \int \frac{u-2}{3\sqrt{u}}. \frac{du}{3} = \frac{1}{9} \left( \int \frac{u}{\sqrt{u}} du - 2\int \frac{1}{\sqrt{u}} du \right) = \frac{1}{9} \left( \int u^{\frac{1}{2}}du -2 \int u^{-\frac{1}{2}} du \right) \)

Calculate the Integral

Now we can easily compute the integral which results in \( \frac{1}{9} ( \frac{2}{3} u^{\frac{3}{2}} - 2(2 u^{\frac{1}{2}}) )+ C = \frac{2}{27}u^{\frac{3}{2}} - \frac{4}{9}u^{\frac{1}{2}} + C \)

Back-Substitution

Do the back-substitution \( u = 2 + 3x \). The final result is \( \frac{2}{27}(2 + 3x)^\frac{3}{2} - \frac{4}{9}(2 + 3x)^\frac{1}{2} + C \)