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5.4. INTEGRATION BY PARTS
algebraic antiderivative. For instance, if we consider the indefinite integrals
e
x 2 dx and
x tan(x) dx,
neither u-substitution nor Integration by Parts proves fruitful. While there are other
integration techniques, some of which we will consider briefly, none of them enables
us to find an algebraic antiderivative for e x 2 or x tan(x). There are at least two key
observations to make: one, we do know from the Second Fundamental Theorem of
Calculus that we can construct an integral antiderivative for each function; and two,
antidifferentiation is much, much harder in general than differentiation. In particular, we
observe that F(x) =
x
0
e t 2 dt is an antiderivative of f (x) = e x 2 , and G(x) =
x
0
t tan(t) dt
is an antiderivative of g(x) = x tan(x). But finding an elementary algebraic formula that
doesn’t involve integrals for either F or G turns out not only to be impossible through
u-substitution or Integration by Parts, but indeed impossible altogether.
Summary
In this section, we encountered the following important ideas:
• Through the method of Integration by Parts, we can evaluate indefinite integrals that
involve products of basic functions such as
x sin(x) dx and
x ln(x) dx through a
substitution that enables us to effectively trade one of the functions in the product for
its derivative, and the other for its antiderivative, in an effort to find a different product
of functions that is easier to integrate.
• If we are given an integral whose algebraic structure we can identify as a product of
basic functions in the form
f (x)g ′ (x) dx, we can use the substitution u = f (x) and
dv = g ′ (x) dx and apply the rule
u dv = uv −
v du
in an effort to evaluate the original integral
f (x)g ′ (x) dx by instead evaluating
v du =
f ′ (x)g(x) dx.
• When deciding to integrate by parts, we normally have a product of functions present
in the integrand and we have to select both u and dv. That selection is guided by the
overall principal that we desire the new integral
v du to not be any more difficult or
complicated than the original integral
u dv. In addition, it is often helpful to recognize
if one of the functions present is much easier to differentiate than antidifferentiate (such
as ln(x)), in which case that function often is best assigned the variable u. For sure,
when choosing dv, the corresponding function must be one that we can antidifferentiate.
5.4. INTEGRATION BY PARTS
algebraic antiderivative. For instance, if we consider the indefinite integrals
e
x 2 dx and
x tan(x) dx,
neither u-substitution nor Integration by Parts proves fruitful. While there are other
integration techniques, some of which we will consider briefly, none of them enables
us to find an algebraic antiderivative for e x 2 or x tan(x). There are at least two key
observations to make: one, we do know from the Second Fundamental Theorem of
Calculus that we can construct an integral antiderivative for each function; and two,
antidifferentiation is much, much harder in general than differentiation. In particular, we
observe that F(x) =
x
0
e t 2 dt is an antiderivative of f (x) = e x 2 , and G(x) =
x
0
t tan(t) dt
is an antiderivative of g(x) = x tan(x). But finding an elementary algebraic formula that
doesn’t involve integrals for either F or G turns out not only to be impossible through
u-substitution or Integration by Parts, but indeed impossible altogether.
Summary
In this section, we encountered the following important ideas:
• Through the method of Integration by Parts, we can evaluate indefinite integrals that
involve products of basic functions such as
x sin(x) dx and
x ln(x) dx through a
substitution that enables us to effectively trade one of the functions in the product for
its derivative, and the other for its antiderivative, in an effort to find a different product
of functions that is easier to integrate.
• If we are given an integral whose algebraic structure we can identify as a product of
basic functions in the form
f (x)g ′ (x) dx, we can use the substitution u = f (x) and
dv = g ′ (x) dx and apply the rule
u dv = uv −
v du
in an effort to evaluate the original integral
f (x)g ′ (x) dx by instead evaluating
v du =
f ′ (x)g(x) dx.
• When deciding to integrate by parts, we normally have a product of functions present
in the integrand and we have to select both u and dv. That selection is guided by the
overall principal that we desire the new integral
v du to not be any more difficult or
complicated than the original integral
u dv. In addition, it is often helpful to recognize
if one of the functions present is much easier to differentiate than antidifferentiate (such
as ln(x)), in which case that function often is best assigned the variable u. For sure,
when choosing dv, the corresponding function must be one that we can antidifferentiate.