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.

Continues below

Example 1


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


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.


Integrate each of the given functions:

Exercise 1

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

Exercise 2


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).