Question

Use the indicated formula from the table of integrals in this section to find the indefinite integral. $$ \int \frac{1}{x(2+3 x)^{2}} d x, \text { Formula } 11 $$

Short answer

So, the indefinite integral \(\int \frac{1}{x(2+3x)^2} dx = \frac{1}{6}\ln\left(\frac{2}{2+3x}\right)+ C\)

Step-by-step solution

Identifying the Corresponding Integral Formula

The formula 11 referred in the problem is the integral rule for rational functions, given by: \[\int \frac{1}{u(a+bu)} du = \frac{1}{ab} \ln \left | \frac{a}{a+bu} \right | + C\]We recognize the integral in the given problem can be rewritten in the form of the formula 11, with \( u = x \), \(a = 2\), and \(b = 3\). This means that the integral \[\int \frac{1}{x(2+3x)^2} dx \]matches with the formula 11 when \( u = x \), \(a = 2\) and \(b = 3\)

Applying the Integral Formula

Substitute the corresponding value from the given integral into the formula, we get: \[\int \frac{1}{x(2+3x)^2} dx = \frac{1}{2*3} \ln \left |\frac{2}{2+3x} \right | + C\]which simplifies to:\[\int \frac{1}{x(2+3x)^2} dx = \frac{1}{6} \ln \left |\frac{2}{2+3x} \right | + C\]

Simplifying the Result

Since the absolute value of a positive real number is the number itself, and the constant 'C' can absorb the absolute value, we can summarize:\[\int \frac{1}{x(2+3x)^2} dx = \frac{1}{6}\ln\left(\frac{2}{2+3x}\right)+ C\]Which is our final answer.