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Chapter 17: Problem 15

Find \(\int \frac{4}{3-x} \mathrm{~d} x\).

Short Answer

Expert verified
Using substitution and integrating, we find out that the integral $$\int \frac{4}{3-x} \mathrm{d}x$$ is equal to the antiderivative $$-4\ln|3-x|+C$$.

Step by step solution

01

Write down the given integral

We are given the integral: $$\int \frac{4}{3-x} \mathrm{~d} x$$
02

Find the antiderivative of \(\frac{4}{3-x}\)

We can use the substitution method to find the antiderivative of \(\frac{4}{3-x}\). Let \(u = 3 - x\). Then, \(\frac{du}{dx} = -1\), so \(dx = -du\). The integral becomes: $$\int \frac{4}{3-x} \mathrm{~d} x = \int \frac{4}{u} (-du) = -4 \int \frac{1}{u} \mathrm{~d} u$$ Now, we know that the antiderivative of \(\frac{1}{u}\) is \(\ln |u|\). Thus, $$-4 \int \frac{1}{u} \mathrm{~d} u = -4 \ln |u| + C$$
03

Substitute back for \(u\) and simplify

Since \(u = 3 - x\), we have $$-4 \ln |u| + C = -4 \ln |3-x| + C$$ Therefore, the solution to the given integral is $$\int \frac{4}{3-x} \mathrm{~d} x = -4 \ln |3-x| + C$$

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