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4.2: Logs and Integrals

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

[egin{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.


Watch the video: Function Equality- Logs and Exponentials (August 2022).