Question

Use the indicated formula from the table of integrals in this section to find the indefinite integral. $$ \int \frac{x}{(2+3 x)^{2}} d x, \text { Formula } 4 $$

Short answer

The indefinite integral \(\int \frac{x}{(2+3x)^2} dx\) evaluates to \(-\frac{x}{3(2+3x)}+\frac{1}{3}\ln|2+3x| + C\).

Step-by-step solution

Identify the Integral Formula

The first step is to recognize the type of the integrand and correspond it with the appropriate formula. As follows: \[\int u dv = uv - \int v du\] This is referred to as 'Formula 4' or the formula for integration by parts.

Identification of Coefficients

We identify the value of \(u, du, v\) and \(dv\). In our integral \(\int \frac{x}{(2+3x)^2} dx\), we can take \(u=x\), so \(dv=\frac{1}{(2+3x)^2}dx\). Therefore, \(du=dx\). For \(v\), we evaluate the integral of \(dv\), that is, \(\int \frac{1}{(2+3x)^2}dx\), which turns out to be \(-\frac{1}{3(2+3x)}\). Note that a substitution method is used to find the integral of \(dv\), by setting \(t=(2+3x)\), and then finding the integral of \(\frac{1}{t^2}\).

Use the Formula to Evaluate the Integral

Using the formula, the integral can be evaluated as: \[\int \frac{x}{(2+3x)^2} dx = uv - \int v du = x \left(-\frac{1}{3(2+3x)}\right) - \int -\frac{1}{3(2+3x)} dx\]Simplify it to: \[-\frac{x}{3(2+3x)}+\frac{1}{3}\int \frac{1}{2+3x} dx\]The integral on the right side can be solved using straightforward integration and the result turns out to be \(\frac{1}{3}\ln|2+3x|\).

Combine the Results

The result from the previous step is now combined to give us the final answer:\[-\frac{x}{3(2+3x)}+\frac{1}{3}\ln|2+3x| + C\]where \(C\) is the constant of integration.