Integration by Parts Exponentials and Logarithms
James K. Peterson
Department of Biological Sciences and Department of Mathematical Sciences
Clemson University
September 17, 2013
Outline
Resources
Integration By Parts
IbyP with exponentials: One Step
IbyP with logarithms
Abstract
This lecture is going to talk about a new method of integration
called Integration by Parts.
This technique is based on the product rule for differentiation.
I
Let’s assume that the functions f and g are both
differentiable on the finite interval [a, b]. Then the product fg
is also differentiable on [a, b] and
(f (t) g (t))0 = f 0 (t) g (t) + f (t) g 0 (t)
I
Now, we know the antiderivative of (fg )0 is fg + C . So, if we
compute the definite integral of both sides of this equation on
[a, b], we find
Z
a
b
(f (t) g (t))0 dt =
Z
a
b
f 0 (t) g (t) dt +
Z
a
b
f (t) g 0 (t) dt
I
The left hand side is simply (fg ) |ba = f (b)g (b) − f (a)g (a).
Hence,
Z
b
f 0 (t) g (t) dt +
a
I
Z
b
f (t) g 0 (t) dt = (fg ) |ba
a
This is traditionally written as
Z
b
f (t) g 0 (t) dt = (fg ) |ba −
a
I
g (t) f 0 (t) dt
a
We usually write this in an even more abbreviated form. If we
let u(t) = f (t), then du = f 0 (t) dt. Also, if v (t) = g (t),
then dv = g 0 (t) dt. We then rewrite as
Z
b
u dv
= uv |ba −
a
I
b
Z
Z
b
v du
a
We can also develop the integration by parts formula as an
indefinite integral. When we do that we obtain the formula
commonly referred to as Integration by Parts – IbyP.
Z
Z
u dv
= uv −
v du + C
I
So given a new integral to do, if we can’t do a simple
substitution, another possibility is IbyP.
I
Then it is all about looking at the new integral and guessing
what part of it should be u and what part of it should be dv.
I
This is best done by examples.
I
We can use IbyP to handle
Z
( some polynomial of t ) e t dt.
Example:
R
(2t 2 + 4)e t dt.
I
We can complicate it aRbit by letting the argument inside the
ln be linear. Example: (2t 2 + 4)e 5t dt.
I
And we can spin out more cases, but easier to see how it plays
out with some examples.
I
These are harder than the ln ones we’ll do next.
Example
Evaluate
R
t e t dt
Solution
t
Let u(t)
R = t and dv = e dt. Then du = dt and
v = e t dt = e t . Applying IbyP, we have
Z
Z
Z
t e t dt =
udv = uv −
vdu
Z
t
t
= e t −
e dt
Z
t
= te −
e t dt = t e t − e t + C
Example
Evaluate
R
(2t + 5) e −4t dt
Solution
Let u(t) = 2t + 5 and dv = e −4t dt. Then du = 2dt and
v = (−1/4)e −4t . Applying IbyP, we have
Z
Z
Z
(2t + 5) e −4t dt =
udv = uv −
vdu
Z
−1 −4t
−1 −4t
=
e
(2t + 5) −
e
2dt
4
4
Z
2
2t + 5 −4t
−4t
= −
e
+
e
dt
4
4
2t + 5 −4t
2 −1 −4t
= −
e
+
e
+C
4
4 4
Example
Evaluate
R
t 2 e t dt
Solution
Step 1:
2
t
Let u(t)
R = t and dv = e dt. Then du = 2tdt and
v = e t dt = e t . Applying Integration by Parts, we have
Z
Z
Z
t 2 e t dt =
udv = uv −
vdu
Z
= et t2 −
e t 2t dt
Z
= t2 et −
2t e t dt
Solution
Step 2:
R
Now the integral 2t e t dt also requires the use of integration by
parts. So we integrate again using this technique
Let u(t) = 2t
R
and dv = e t dt. Then du = 2dt and v = e t dt = e t . Applying
IbyP again, we have
Z
Z
2t e t dt =
udv
Z
= uv −
vdu
Z
t
= e 2t −
e t 2 dt
Z
= 2t e t −
2 e t dt
= 2t e t − 2 e t + C
Solution
Step 3:
Assemble the whole shebang:
Z
Z
t 2 e t dt = t 2 e t −
2t e t dt
= t 2 e t − 2t e t − 2 e t + C
It is very awkward to do these multiple integration by parts in two
separate steps like we just did. It is much more convenient to
repackage the computation like this:
Solution
R
t 2 e t dt
= uv −
t2
e t dt
u =
dv =
du = 2tdt v = e t
R
vdu
R t
=
−
e 2t dt
R
= e t t 2 − R e t 2t dt
= t2 et −
2t e t dt
et
t2
u = 2t
dv = e t dt
du = 2dt v = e t
R
= t 2 e t − e t 2t − e t 2 dt
= t 2 e t − {2t e t − 2 e t } + C
=
This is still awkward (and long!) for our problem solution. So let’s
try this:
Solution
Z
t 2 e t dt =
u = t 2 ; dv = e t dt; du = 2tdt; v = e t
Z
= et t2 −
e t 2t dt
u = 2t; dv = e t dt; du = 2dt; v = e t
Z
= t e − e 2t −
e t 2 dt
= t 2 e t − 2t e t − 2 e t + C
2
t
t
= t 2 e t − 2t e t + 2 e t + C
Example
Evaluate
R
(t 2 + 4t + 5) e −4t dt
Solution
Z
(t 2 + 4t + 5) e −4t dt =
u = t 2 + 4t + 5; dv = e −4t dt
−1 −4t
e
4
Z
−1 −4t
−1 −4t
2
= (t + 4t + 5) e
−
e
(2t + 4) dt
4
4
Z
t 2 + 4t + 5 −4t 1
=−
e
+
e −4t (2t + 4) dt
4
4
du = (2t + 4)dt; v =
Solution
u = 2t + 4; dv = e −4t dt;
−1 −4t
du = 2dt; v =
e
4
Z
t 2 + 4t + 5 −4t 1 −1 −4t
−1 −4t
= −
e
+
e
(2t + 4) −
e 2dt
4
4
4
4
Z
t 2 + 4t + 5 −4t 1 −1 −4t
2
= −
e
+
e
(2t + 4) +
e −4t dt
4
4
4
4
t 2 + 4t + 5 −4t 1 −1 −4t
2 −1 −4t
= −
e
+
e
(2t + 4) +
e
+C
4
4
4
4 4
Homework 25
t 2 e 3t dt
25.1 Evaluate
R
25.2 Evaluate
R2
(6t + 8) e −2t dt
25.3 Evaluate
R0
(5t + 3) e 2t dt
25.4 Evaluate
R
(5t 2 + 13) e 4t dt
25.5 Evaluate
R
(7t − 3) e −5t dt
I
We can use IbyP to handle
Z
( some polynomial of t ) ln(t) dt.
Example:
R
(2t 2 + 4) ln(t)dt.
I
We can complicate it aRbit by letting the argument inside the
ln be linear. Example: (2t 2 + 4) ln(5t)dt.
I
And we can spin out more cases, but easier to see how it plays
out with some examples.
Example
Evaluate
R
ln(t) dt
Solution
1
Let u(t)
R = ln(t) and dv = dt. Then du = t dt and
v =
dt = t. For the antiderivative v don’t add an arbitrary
constant C as we will add one at the end. Applying IbyP,
Z
Z
Z
ln(t) dt =
udv = uv −
vdu
Z
1
= ln(t) t −
t dt
t
Z
= ln(t) t −
dt = t ln(t) − t + C
Example
Evaluate
R
t ln(3t) dt
Solution
3
1
Let u(t)
R = ln(3t) and dv = tdt. Then du = 3t dt = t dt and
v = tdt = t 2 /2. Using IbyP
Z
Z
Z
t ln(3t) dt =
udv = uv −
vdu
Z 2
t 1
2
= ln(3t) t /2 −
dt
2 t
Z
t2
=
ln(3t) −
t/2 dt
2
2
2
t
t
=
ln(3t) −
+ C
2
4
Example
Evaluate
R
t 3 ln(8t) dt
Solution
8
1
3
Let u(t)
R = ln(8t) and dv = t dt. Then du = 8t dt = t dt and
v = t 3 dt = t 4 /4. Applying Integration by Parts, we have
Z
Z
Z
t 3 ln(8t) dt =
udv = uv −
vdu
Z
1
4
4
= ln(8t) t /4 −
t /4 dt
t
Z
4
t
=
ln(8t) −
t 3 /4 dt
4
t2
t4
=
ln(8t) −
+ C
2
16
Example
Evaluate
R
(t 4 + 5t 2 + 8t + 9) ln(4t) dt
Solution
Let u(t) = ln(4t) and dv = (t 4 + 5t 2 + 8t + 9)dt. Then
du =R4t4 dt = 1t dt and
v = (t 4 + 5t 2 + 8t + 9)dt = t 5 /5 + (5/3)t 3 + (8/2)t 2 + 9t. Applying
Integration by Parts, we have
Z
(t 4 + 5t 2 + 8t + 9) ln(4t) dt = ln(4t)(t 5 /5 + (5/3)t 3 + (8/2)t 2 + 9t)
Z
1
− (t 5 /5 + (5/3)t 3 + (8/2)t 2 + 9t) dt
t
= ln(4t)(t 5 /5 + (5/3)t 3 + (8/2)t 2 + 9t)
Z
− (t 4 /5 + (5/3)t 2 + (8/2)t + 9) dt
= ln(4t)(t 5 /5 + (5/3)t 3 + (8/2)t 2 + 9t)
−(t 5 /25 + (5/9)t 3 + (8/4)t 2 + 9t) + C
Homework 26
26.1 Evaluate
R
26.2 Evaluate
R
2t ln(t 2 ) dt
26.3 Evaluate
R
(t 4 + 5t − 8) ln(7t) dt
26.4 Evaluate
R5
2
ln(5t) dt
(t 2 + 5t + 3) ln(t) dt