# 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

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

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

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.

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)

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

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.

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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