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# Integration by Parts

Unlike differentiating, there is no systematic procedure for finding
antiderivatives. The rule of substitution discussed in the previous section is
effective if we can find a substitution u (x) for which a term u '(x) appears in
the integral. Substitution may not work, but another powerful technique is
available based on the product rule for derivatives. Integration by parts is
given by the following formula.

To see this, note that

from the product rule for derivatives. Also,

this statement about antiderivatives is true simply because the derivative of

Consequently,

Thus,

Since the antiderivative on the right side already involves a constant of
integration,c is redundant and can be omitted. This gives us the formula for
integrating by parts.

Remark: The g (x) in the integral on the right side of Formula (1) is any
antiderivative of g '(x) in the integral on the left side; therefore, the
constant in g (x) on the right side may be chosen to be any convenient value. It
is usually chosen to be 0, although occasionally a different choice is useful.

**Example 1** Find

To write this integral in the form

in order to apply Formula (1), we must determine how to choose f and g. The
rule of thumb is to choose for g '(x) the "most difficult part" of the integrand
which can be integrated. In this example, the rule suggests that we let g'(x) =
e^{x}, and thus f(x) = x. With this choice,

Clearly, f'(x) = I and g(x) = e^{x}. (Since g(x) is any
antiderivative, we choose c = 0 in the general antiderivative e^{x} + c
.) That is,

Hence,

**Example 2** Find

We have a choice to make: either let f(x) = ln(x) and g'(x) = 1 or let f(x) = 1
and g'(x) = ln(x). If we choose g'(x) = ln(x) and f(x) = l, we must then find
g(x) by integrating ln(x), which is, unfortunately, exactly our problem in this
example.

Hence, let f(x) = ln(x) and g'(x) = 1. Then f '(x) = 1/x and g(x) = x. We now
have

**Example 3** Find

We could let g '(x) = ln(x) and use the result of Example 2, but then g (x)
would be fairly complicated. Instead we let f(x) = ln(x) and g'(x) = x^{1/2},
in which case f '(x) = l/x and g(x) = 2/3x^{3/2}.Hence,

The next example illustrates how we can use integration by parts to evaluate a
definite integral. The fundamental theorem of calculus is used together with the
fact that the integration by parts formula changes one antiderivative into
another. The integration by parts formula is

Hence, the corresponding formula for the definite integral is the following:

**Example 4 **Find

We begin by using the conventional techniques for integrating by parts. As in
Example 1, let f(x) = x and g'(x) = e^{x}. Then f '(x) = 1 and g(x) = e^{x},
and hence, by Formula (3)