Chapter 8
Methods for Finding AntiderivativesThere are methods for finding antiderivatives. These standard methods are recipes. Theydon’t always work but they are the best we have. It turns out you can’t find antiderivativesin terms of elementary functions in a routine way as you can with finding derivatives. It isa skill, not a substantial part of mathematics.
8.1 The Method of SubstitutionI will illustrate the method of substitution by the use of examples. The method is somewhatformal. However, it works and you can check the answers obtained. Ultimately it is basedon the chain rule for derivatives.
Example 8.1.1 Find∫ 3√
2x+7xdx.
In this example u = 2x+7 so that du = 2dx. Then
∫3√2x+7xdx =
∫3√
u
x︷ ︸︸ ︷u−7
2
dx︷︸︸︷12
du =∫ (1
4u4/3 − 7
4u1/3
)du
=328
u7/3 − 2116
u4/3 +C =3
28(2x+7)7/3 − 21
16(2x+7)4/3 +C
Example 8.1.2 Find∫
xex2dx.
Define a new variable u = x2. Then dudx = 2x and so du = 2xdx and xdx = 1
2 du. Thenin terms of u the above integral is 1
2∫
eudu = 12 eu +C. Now substitute in what u equals in
terms of x. This yields 12 ex2
+C. Next check your work. Take the derivative of what youthink the answer is and verify that it really is an antiderivative.
Example 8.1.3∫
sin(x)cos(x)√
1+ sin2 (x)dx
This can be done as follows. Let u = 1+ sin2 (x) so du = 2sin(x)cos(x)dx and theintegral in the example simplifies to
12
∫ √udu =
13
u32 +C =
13(1+ sin2 (x)
) 32 +C
You might try letting u = sin2 (x). It will also work but will likely take longer.This illustrates that you are not always sure what substitution to use, but ultimately this
method depends on the chain rule.∫f (g(x))g′ (x) dx = F (g(x))+C, (8.1)
where F ′ (y) = f (y). Here is another example.
Example 8.1.4 Find∫
x3x2dx
191