Question

this is similar to section 4.2 problem 30:
determine the indefinite integral. use capital c for the free constant.
$$ int ( \frac { - 6 } { x ^ { 3 } } - \frac { 4 } { x } ) d x = $$
hint: follow example 9.
resources
ebook

Explanation:

Step1: Split the integral

$$\int(\frac{-6}{x^{3}}-\frac{4}{x})dx=\int\frac{-6}{x^{3}}dx-\int\frac{4}{x}dx$$

Step2: Rewrite the first - term

$$\int\frac{-6}{x^{3}}dx=-6\int x^{-3}dx$$
Using the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), when \(n=-3\), we have \(-6\times\frac{x^{-3 + 1}}{-3+1}=-6\times\frac{x^{-2}}{-2}=3x^{-2}\)

Step3: Integrate the second - term

$$\int\frac{4}{x}dx = 4\int\frac{1}{x}dx$$
Using the formula \(\int\frac{1}{x}dx=\ln|x|+C\), we get \(4\ln|x|\)

Answer:

\(3x^{-2}-4\ln|x|+C\)