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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`

Answer

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

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

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

`=1/2ln|u|+K`

`=1/2ln|x^4+1|+K`

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`

Answer

`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`

`=0.223`

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`

Answer

`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

`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.

Answer

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

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

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

`=-int(du)/u`

`=-ln|u|+K`

`=-ln|20-v|+K`

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:

`t=ln(20/(20-v))`

Taking "`e` to both sides":

`e^t=20/(20-v)`

Inverting the fraction gives:

`e^-t=(20-v)/20`

`20e^-t=20-v`

` v=20-20e^-t`

`v=20(1-e^-t)`

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)`

Answer

`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

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

Answer

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

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

`int_1^2(x^2+1)/(x^3+3x)dx=1/3int_(x=1)^(x=2)(du)/u`

`=1/3[ln|u|]_(x=1)^(x=2)`

`=1/3[ln|x^3+3x|]_1^2`

`=1/3[ln(14)-ln(4)]`

`=0.418`

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.

Answer

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

Put `u = 2 + cos π t`, then

`du = −π\ sin π t\ dt`

So

`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).

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