The Integration Power Rule

Of all the integration rules, this is the most used and , the most important to understand. This rule takes the form of ∫uⁿdu = uⁿ⁺¹ /(n+1) + C, for any power where n not equal to -1. In that case, The integral becomes ∫uⁿdu = ln|u|+C, a very special form, ln is the natural log.

The tricky part of the power rule is making sure all of du is present before we integrate. If it is not present in its entirety, we need to make an adjustment to make sure we have it. However, the only type of adjustment that can be made is by multiplying the inside of the integral by a Constant then dividing the outside by the same constant. That way, the integral is not altered and the answer is not changed. This seems to be difficult for most students to understand. Any type of variable adjustment above is not allowed.

Once the correct adjustment is made (assuming it can be made), we can proceed to the answer given by the rule. Note that in the answer the du is absent, it simply gets absorbed back into the expression.

In the cases where a constant adjustment can not be made, the indefinite integral can not be found. Fortunately, most integrals of this type are definite integrals (i.e., limits on the integral symbol) and can be approximated very nicely by our calculator. (these are just numbers).

In the case where u=x+a, where a is any constant, du=dx, and no adjustment is necessary since dx is the correct du automatically. We called these forms the simple cases of the power rule.

Examples: 1) ∫(7x-5)³dx. Since u=7x-5, du=7dx (the differential of an expression is the derivative times the differential of the variable of integration). So, we need to multiply the inside by 7 and divide the outside by 7 (or multiply the outside by 1/7). We then have, (1/7)∫(7x-5)³ 7dx =(1/7)(7x-5)⁴/4 +C, or (1/21)(7x-5)⁴ +C

Note that the du=7dx is absent in the answer. (it is absorbed back into the problem).

2) ∫x⁵3dx. Since the u=x, the du=dx and no adjustment is necessary. However, we need to eliminate the 3 from the inside. So, simply move it to the outside (constants have free passage). We get, 3∫x⁵dx = (3/6)x⁶+C or (1/2)x⁶+C. This is an example of the simply case.

3) Be aware that many problems are “rigged up” so that the du can be present. For example, take ∫(2x²+3x+4)⁸(6+8x)dx. Note that u=2x²+3x+4 and du=(4x+3)dx. By factoring out a 2 and rearranging the part in front of dx, we have, 2∫(2x²+3x+4)⁸(4x+3)dx. The integral is now in perfect form. The answer is (2/9(2x²+3x+4)⁹ +C. Note that u is unchanged thoughout this process and the du part disappears.