1.2.3 Integrals Involving Exponentials and logarithms.
In this section we look at integrals involving exponentials and logarithms. From the derivative formulas of
exponentials and logarithms we obtain the following integral formulas.
Proposition 1.
1
(1)

 x dx = ln x
(2)

 x dx = ln | x |
(3)
x
x

 e dx = e
(4)
 ax dx = ln a

if x > 0
1
if x  0
ax
Proof. Formulas (1) and (3) follow from the differentiation formulas in Proposition 1 in Section 1.2.2.
d
Formula (4) follows from the differentiation formula [ aex ] = (ln a) ax. Formula (2) is the same as
dx
formula (1) for x > 0. To prove it for x < 0, note that | x | = - x for x < 0. So if x < 0 one has
d
d
1 d
1
1
ln | x | = ln(- x) =
(- x) =
(-1) = and (2) follows from this. \\
dx
dx
- x dx
-x
x
2
-(x )
Example 1. Find 
 x e dx.
2
1
First we ignore the limits and do the indefinite integral 
 x e-(x ) dx. Make the substitution u = - x2. Then
2
du
1
1
1
1 u
u
 x e-(x2) dx =  eu - 2 du = - 2 
= - 2x and du = - 2x dx and - du = x dx. So 
 e du = - 2 e =
dx
2

2
1 2
-(x2)
- e-(x ). To evaluate the definite integral 
 x e dx we plug in the limits and subtract.
2
1
e
-(x2)
2
1
-(x2)

 x e dx = - 2
1
1 -(22) 1 -(12) e-1 - e-4 0.368 - 0.018 0.350
=
e
+ e
=


 0.175. Alternatively, we could have also
1
2
2
2
2
2
2

2
-4
-(x )
substituted the limits when we made the substitution so that 
 x e dx = 2
1
 0.175.
1.2.3 - 1
1
1 u
u
 e du = - 2 e
2
-1
-4

-1
=
e-1 - e-4
2
Example 2. Find 

1
1
dx. Make the substitution u = 2x - 1. Then du = 2dx and du = dx. So
2x - 1
2
 1 dx = 1  1 du = 1 ln u = 1 ln(2x – 1). If 2x – 1 could be negative, then we would write the answer
2u
2
2
 2x - 1
as
1
ln| 2x – 1 |.
2
Problem 1. Do the following integrals.
a.
2x

 e dx
b.
t

 5 dt
c.
x
x 10

 e (1 + e ) dx
d.

 tan x dx
e.
 ex sin(ex) dx

f.


h.

 4 - x3 dx
i.

x2
-2
g.
1

 x dx
(1 + ln x)4
dx
x
x2
-5
i.
1

 ex dx
2
1
1.1 - 2
-(x2)
dx.