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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: $$\int(9x - 2)^{-3} dx$$ Answer: $$\int(9x - 2)^{-3} dx = \frac{-1}{18}(9x-2)^{-2} + \frac{C}{9}$$

Step by step solution

01

Identify the substitution variable u

Let's identify what part of the integral can be replaced by a new variable u. In this case, we choose $$u = (9x - 2)$$ because it is the base of the exponent in the integrand.
02

Differentiate u with respect to x

Now, we need to find the differential of u with respect to x, which is $$\frac{d u}{d x}$$. Differentiating u with respect to x, we get: $$\frac{d u}{d x} = \frac{d (9x - 2)}{d x} = 9$$
03

Express dx in terms of du

We will now express \(dx\) in terms of \(du\). In order to do that, we need to rewrite \(\frac{du}{dx}\) as \(\frac{dx}{du}\): $$\frac{dx}{du} = \frac{1}{9}$$ Now, we can obtain the expression for \(dx\) in terms of \(du\) by multiplying both sides by \(du\): $$dx = \frac{1}{9} du$$
04

Substitute u and dx in the integral and solve

Now it's time to substitute both our new variable u and the expression for dx into the integral: $$\int(9x - 2)^{-3} dx = \int u^{-3} \frac{1}{9} du$$ Let's simplify the integral by taking the constant 1/9 out from the integral: $$\frac{1}{9} \int u^{-3} du$$ Now, we integrate u^{-3} with respect to u: $$\frac{1}{9} \int u^{-3} du = \frac{1}{9} \left[ \frac{u^{-2}}{-2} + C \right]$$ Here, C is the constant of integration.
05

Substitute back the original variable x

Finally, we will substitute back the original variable x into the expression by replacing u with \((9x-2)\): $$\frac{1}{9} \left[ \frac{(9x-2)^{-2}}{-2} + C \right] = \frac{-1}{18}(9x-2)^{-2} + \frac{C}{9}$$ The result of the integral is: $$\int(9x - 2)^{-3} dx = \frac{-1}{18}(9x-2)^{-2} + \frac{C}{9}$$

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