The General Power Rule

In the theory-page Finding Antiderivatives, we used the Simple Power Rule

∫ x n d x = x n + 1 n + 1 + C , n &neq; - 1

to find antiderivatives of functions expressed as powers of x alone. In this theory page, we look at a techniques for finding antideriva-tives for more general functions.

To begin, consider how you might find the following integral.

∫ 2 x ( x 2 + 1 ) 3 d x

You could expand the integrand into polynomial form and the use the Simple Power Rule on the individual power of x. But there is a simpler way to evaluate this integral! To see it, you must remember that we are hunting for a function f such that

f ′ ( x ) = 2 x ( x 2 + 1 ) 3

After some experimentation you might observe that

d d x [ ( x 2 + 1 ) 4 ] = 4 ( x 2 + 1 ) 3 . ( 2 x )

and that by dividing by 4 you could obtain

d d x [ ( x 2 + 1 ) 4 4 ] = ( x 2 + 1 ) 3 . ( 2 x )

Now, having found a function with the desired derivative, you could conclude that

∫ 2 x ( x 2 + 1 ) 3 d x = ( x 2 + 1 ) 4 4 + C

Be sure you see that the key to this solution is the presence of the factor 2x in the integrand. Notice that the factor of the integrand, 2x, is precisely the derivative of x 2 + 1 . That is, if

u = x 2 + 1

then

d u d x = 2 x

Thus, written terms of u, the original integral becomes

∫ 2x ⏟ du dx ( x 2 +1) ⏞ u d x = ∫ u 3 du dx d x = u 4 4 + C

This is just one example of an important integration rule called the General Power Rule.

General Power Rule for Integration

If u is a differentiable function of x, then

∫ u n d u d x d x = u n + 1 n + 1 + C , n &neq; - 1

An important consideration that is often overlooked in using General Power Rule is the existence of du/dx as a factor of the integrand. We must first determine u by identifying the integrand a function u that is raised to a power. Then we must show that its derivative du/dx is also a factor of the integrand. Simply stated, these clues for using the General Power Rule for integration are as follows.

1. Identify the function u that is raised to a power.

2. Check to see if du/dx is also a factor of the integrand.

We demonstrade these steps in the next examples.

EXAMPLE 1
Use the General Power Rule to finding the following indefinite integrals.

∫ 3 ( 3 x - 1 ) 4 d x

SOLUTION
Letting u = 3x-1, we have du / dx = 3.

∫ 3 ( 3 x - 1 ) 4 d x = ∫ ( 3 x - 1 ) 4 ( 3 ) d x = ( 3 x - 1 ) 5 5 + C

EXAMPLE 2
Use the General Power Rule to finding the following indefinite integrals.

∫ ( 2 x + 1 ) ( x 2 + x ) d x

SOLUTION
Letting u = x 2 + x , we have d u / d x = 2 x + 1 .

∫ ( 2 x + 1 ) ( x 2 + x ) d x	= ∫ ( 2 x + 1 ) ( x 2 + x ) d x =
	= ∫ ( x 2 + x ) 1 ( 2 x + 1 ) d x =
	= ( x 2 + x ) 2 2 + C

EXAMPLE 3
Use the General Power Rule to finding the following indefinite integrals.

∫ 3 x 2 x 3 - 2 d x

SOLUTION
Letting u = x 3 - 2 , we have d u / d x = 3 x 2

∫ 3 x 2 x 3 - 2 d x	= ∫ 3 x 2 x 3 - 2 d x = ∫ ( x 3 - 2 ) 1 / 2 ( 3 x 2 ) d x =
	= ( x 3 - 2 ) 3 / 2 3 / 2 + C =
	= 2 3 ( x 3 - 2 ) 3 / 2 + C

REMARK

Example 2 illustrates a case of a General Power Rule that is sometimes overlooked - when the power is n = 1. In this case the rule takes the form

∫ u d u d x d x = u 2 2 + C

Many times part of the derivative du/dx is missing from the integrand and in some cases we can make the necessary adjustments to apply the Genaral Power Rule. In other cases we cannot adjust appropriately and therefore cannot apply the General Power Rule.

You can look for the more examples and test your ability for solving these problems.