Problem 1 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

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 1: 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.

Step 2: 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 \(\frac{1}{t^2}\).

Step 3: 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 \(\frac{1}{3}\ln|2+3x|\).

Step 4: 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.