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Chapter 7: Problem 8

Evaluate the following integrals. $$\int(9 x-2)^{-3} d x$$

Short Answer

Expert verified
Question: Evaluate the integral of the given function: $$\int (9x-2)^{-3} dx$$ Answer: The integral of the given function is $$\int(9x-2)^{-3} dx = \frac{-1}{18}(9x-2)^{-2} + C$$

Step by step solution

01

Identify Substitution

Since the given integral is similar to the integral of a power function multiplied by the derivative of its inside function, we will apply substitution method. Let's start by choosing a substitution value and its derivative. Let \(u = 9x - 2\). Now, find the derivative with respect to x, $$\frac{d u}{d x} = 9.$$ Now the function is simpler to work with.
02

Substitute and simplify the integral

Express the integral in terms of \(u\). Replace \((9x-2)\) with \(u\) and multiply by the reciprocal of the \(du/dx=9\). $$\int (9x-2)^{-3} dx = \int u^{-3} \cdot \frac{1}{9} du$$ Now, we can integrate the function with respect to \(u\). $$= \frac{1}{9} \int u^{-3} du$$
03

Integrate with respect to \(u\)

Apply the power rule for integration, which states that: $$\int u^n du = \frac{u^{n+1}}{n+1} + C$$ In our case, \(n=-3\), so we have: $$= \frac{1}{9} \cdot \frac{u^{-3+1}}{-3+1} + C$$ Simplify the expression: $$= \frac{1}{9} \cdot \frac{u^{-2}}{-2} + C$$
04

Substitute back and simplify the answer.

Replace \(u\) with \(9x-2\) and simplify the expression: $$= \frac{1}{9} \cdot \frac{(9x-2)^{-2}}{-2} + C$$ Now, multiply: $$= \frac{-1}{18}(9x-2)^{-2} + C$$ The final answer is: $$\int(9x-2)^{-3} dx = \frac{-1}{18}(9x-2)^{-2} + C$$

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