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# 4.2: Logs and Integrals

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Recall that

$\int \dfrac{1}{x} dx = \ln |x| + C.$

Note that we have the absolute value sign since for negative values of that graph of $$\frac{1}{x}$$ is still continuous.

$\int \dfrac{dx}{1-3x}.$

Solution

Let $$u = 1-3x$$ and $$du = -3\, dx$$.

The integral becomes

\begin{align} -\dfrac{1}{3} \int \dfrac{du}{u} &= \dfrac{1}{3}\ln |u| +C \\ &= -\dfrac{1}{3} \ln |1-3x| +C. \end{align}

15) $$\csc x$$ (hint: Use the formula $$\csc x = \sec (\pi/2 - x)$$.

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.