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.