MAC2312
Quiz 3a
Name
Z
1. Evaluate
x sec x tan x dx
Answer. Integrate by parts, with
u=x
dv = sec x tan x dx
du = 1 v = sec x
Then we get
Z
Z
x sec x tan x dx = x sec x −
sec x dx
= x sec x − ln | sec x + tan x|
Answer: x sec x − ln | sec x + tan x
2. True or False?
a) The polynomial 2x4 − 3x3 − 19x2 − 6x + 8 is divisible by (x + 1)
(being divisible means there is 0 remainder after division).
Answer: True
b) The following is a correct first step for breaking up a fraction:
A
B
C
Dx + E
1
=
+
+
+
.
(x − 1)2 (x2 + 1)(x2 − 2x + 12) x − 1 (x − 1)2 x2 + 1 x2 − 2x + 12
Answer: False
Z
3. Evaluate
ex dx
e2x − 1
Answer. Set u = ex , and so du = ex dx. Then the integral becomes
Z
du
=
x
(e )2 − 1
Z
du
=
2
u −1
Z
du
(u + 1)(u − 1)
Now we use partial fractions to break up this fraction:
A
B
1
=
+
(u + 1)(u − 1)
u+1 u−1
1 = (u − 1)A + (u + 1)B
Plugging in u = 1, u = −1 gives A = −1/2, B = 1/2.
So now, the problem is
Z
Z
du
du
−1/2
+ 1/2
u+1
u−1
which equals
u − 1
.
−1/2 ln |u + 1| + 1/2 ln |u − 1| = 1/2 ln u + 1
Substituting back in for u gives
r
x
x
e − 1
= ln e − 1 .
1/2 ln x
e + 1
ex + 1
Answer: ln
q
ex −1
ex +1