# 2. Integration: The Basic Logarithmic Form

by M. Bourne

The general power formula that we saw in Section 1 is valid for all values of *n* **except** *n* = −1.

If *n* = −1, we need to take the opposite of the derivative of the logarithmic function to solve such
cases:

`int(du)/u=ln\ |u|+K`

The `|\ |` (absolute value) signs around
the *u* are necessary since the log of a negative number is
not defined. If you need a reminder, see absolute value.

We can also write the formula as:

`int(f^'(x))/(f(x))dx=ln\ |f(x)|+K`

In words, this means that if we have the derivative of a function in the numerator (top) of a fraction, and the function in the denominator (bottom) of the fraction, then the integral of the fraction will be the natural logarithm of the function.

### Example 1

`int(2x^3)/(x^4+1)dx`

### Example 2

`int_0^(pi//4)(sec^2x)/(4+tan x)dx`

Here is the curve `y=(sec^2x)/(4+tan x)`:

The shaded region represents the integral we just found.

### Example 3

`int(dx)/(x\ ln\ x`

### Example 4

The equation

`t=int(dv)/(20-v)`

comes
from considering a force proportional to the velocity of an
object moving down an inclined plane. Find the velocity,
*v*, as a function of time,
*t*, if the object starts from rest.

This is the graph of the velocity of the sliding object at time *t*.

## Exercises

Integrate each of the given functions:

### Exercise 1

`int(dx)/(x(1+2\ ln\ x)`

### Exercise 2

`int_1^2(x^2+1)/(x^3+3x)dx`

Here's the graph of `y=(x^2+1)/(x^3+3x)`:

The shaded region is the integral we just found.

### Exercise 3

The electric power *p* developed in a
certain resistor is given by

`p=3int(sin\ pi t)/(2+cos\ pi t)dt`

where *t *is the time. Express
*p* as a function of
*t*.

Here's the graph of our solution:

The graph of the power *p* at time *t* (using *K* = 2).

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