Skip to main content
Search IntMath

450+ Math Lessons written by Math Professors and Teachers

5 Million+ Students Helped Each Year

1200+ Articles Written by Math Educators and Enthusiasts

Simplifying and Teaching Math for Over 23 Years

Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.

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




Let `u=x^4+1`, then `du=4x^3dx`.




In this example, we don't actually need the absolute value signs because `x^4+1>0` for all real values of `x`.

Example 2

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


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

Let `u=4+tan x`, then `du=sec^2x\ dx`

`int_0^(pi//4)(sec^2x)/(4+tan x)dx =[ln\ |4+tan x|]_0^(pi//4)`

`=[ln\ |4+tan (pi/4)|]-` `[ln\ |4+tan 0|]`

`=ln\ 5-ln\ 4`


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`


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

Let `u=ln\ x`, then `du=1/xdx`

`int(dx)/(x\ ln\ x)=int(du)/u`

`=ln\ |u|+K`

`=ln\ |ln\ x|+K`

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.



Let `u=20-v`, then `du=-dv`

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




When `t=0`, `v=0`

So `0=-ln\ 20+K`

Therefore `K=ln\ 20`

So `t=ln\ 20-ln(20-v)`

Applying the log laws, we get:


Taking "`e` to both sides":


Inverting the fraction gives:



` v=20-20e^-t`


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)`


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

Put `u = 1 + 2\ ln\ x`, then `du=2/xdx`.

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

`=1/2ln\ |1+2\ ln\ x|+K`

Exercise 2




Put `u = x^3+ 3x`, then `du = (3x^2+ 3) = 3(x^2+ 1)\ 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.


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

Put `u = 2 + cos π t`, then

`du = −π\ sin π t\ dt`


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

`=-3/piint(-pi\ sin pi t)/(2+cos pi t)dt`

`=-3/pi[ln|2+cos pi t|]+K`

Here's the graph of our solution:

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

Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.