Honors Finite – Sec. 12.1 – Integration by Parts To integrate the product of two functions where neither function is the derivative or can be made into the derivative of the other by multiplying by a constant, use integration by parts. Integration by Parts For differentiable functions, u and v, udv uv vdu We substitute u for part of the given integral and dv for the rest, and then express the integral in the form uv vdu . The point is to choose the u and the dv so that the resulting integral vdu is simpler than the original integral udv . Example 1: xe 3x dx Guidelines for choosing u and dv: 1. Choose dv to be the most complicated part of the integral that can be integrated easily. 2. Choose u so that u ' is simpler than u. Example 2: ( x 1)( x 1) 3 Examine the problem to see if integration by parts is necessary. Example 3: Which of the following integral require integration by parts, and which can be found by the substitution formula eu du eu C ? a. xe dx x 2 b. xe x dx Honors Finite – 12.2 – Integration Using Tables We are studying only some of the integration techniques. There are many more advanced techniques which lead to integration formulas. A short table of integrals are on the inside back cover of our book. We look at the problem and try to find a formula that fits it. Example 1: x 2 1 Hint: Use formula 15 with a=5. 25 Sometimes a substitution is needed to transform a formula to fit a given integral. Find a formula that matches the most complicated part of the integral, making appropriate substitutions to change the formula into the given integral. Example 2: x x 1 4 dx Hint: Use formula 18 and use the substitution u x 2 . Sometimes a formula must be used several times to simplify an integral in stages. See example 6 on page 882.