Math 1A - ICA #6 (11-30-16) Fall 2016 Section 3231
Instructor: John Kwon
In-Class Assignment #6: Tricky Integrals - Solution
INSTRUCTION: Do not use a calculator.
Z
1.
dx
x ln x ln (ln x)
Solution. We use u-substitution: Let u = ln x, so that du =
Z
dx
=
x ln x ln (ln x)
Z
1
dx . Then
x
du
.
u ln u
Now use the substitution again, this time with w = ln u, so that dw =
Z
du
=
u ln u
Z
1
du . Then
u
dw
= ln |w| + C = ln |ln u| + C = ln |ln(ln x)| + C .
w
Z
2.
0
ln
√
2
3
e2x
dx
e4x + 1
Solution. We use u-substitution. First, note that the given integral can be written as
Z
ln
√
3
2
0
e2x
1
dx =
4x
e +1
2
ln
Z
√
3
2e2x
2
(e2x )2 + 1
0
dx
Let u = e2x , so that du = 2e2x dx. Then
u = e2·0 = 1
For x = 0 :
For x =
ln
√
2
3
√
3
2· ln 2
:
u=e
= eln
√
3
=
√
3
so the integral is calculated as follows:
Z
0
ln
√
2
3
√
Z √3
1
du
dx =
2
2
2 1 u +1
(e2x ) + 1
0
√3
i
1
1 h −1 √
=
tan−1 u =
tan
3 − tan−1 1
2
2
1
h
i
1 π π
1 π
π
=
−
=
·
=
.
2 3
4
2 12
24
e2x
1
dx =
4x
e +1
2
Z
ln
2
3
2e2x