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# 4.2: Logs and Integrals

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.

Example 1

Evaluate the integral

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

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}

Exercises

Evaluate the integrals of the following:

1)  $$\dfrac{1}{(x-1)}$$

2)  $$\dfrac{1}{(1-x)}$$

3)  $$\cot x$$

4)  $$\dfrac{(2x - 1)}{(x + 2)}$$

5)  $$\dfrac{3x}{(x^2 + 1)^2}$$

6)  $$\dfrac{1}{x \ln x}$$

7)  $$\dfrac{1}{\sqrt{x - 1}}$$

8)  $$\dfrac{(x^2 + 2x + 4)}{(3x)}$$

9)  $$\dfrac{(x + 1)}{(x^2 + 2x)^3}$$

10)  $$(4 - x)^5$$

11) $$\dfrac{1}{\sqrt{3x}}$$

12) $$\tan x$$

13) $$(\tan x)(\ln(\cos x))$$

14)  $$\sec x$$  (hint:  multiply top and bottom by $$\sec x + \tan x)$$

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

### Contributors

• Integrated by Justin Marshall.