Engineering Math II – Spring 2015
Problem 1 (5 pts). The partial fraction decomposition of
(a)
A
x−6
+
B
x+1
+
C
x−1
+
Dx+E
x2 +3
(b)
A
x−6
+
B
x+1
+
C
x−1
+
Dx+E
(x−1)2
(c)
A
x2 −5x−6
(d)
(e)
A
x−6
+
Ax+B
x2 −5x−6
+
B
x−1
+
C
(x−1)2
B
x+1
+
C
x−1
+
+
C
x−1
+
D
x2 +3
+
D
(x−1)2
+
Quiz #4 Solutions
x+2
(x2 −5x−6)(x−1)2 (x2 +3)
is
F x+G
x2 +3
D
x2 +3
+
Ex+F
x2 +3
Solution. The correct answer is (d). The first step is to check that the degree of the
numerator is at most the degree of the denominator. This is clearly the case, since x + 2
has degree 1, and the denominator has degree 6 (add the degrees of each factor in the
denominator). Next, we factor the denominator completely. Note that x2 − 5x − 6 =
2
(x − 6)(x + 1), x − 1 cannot be factored
√ since it already has degree 1, and x + 3 cannot
be factored since its discriminant is −12, which is imaginary. Thus the denominator
factors into (x − 6)(x + 1)(x − 1)2 (x2 + 3).
Now we can write down the partial fraction decomposition (PFD). Since x − 6 and
B
A
and x+1
. The factor x − 1 is
x + 1 are linear, the corresponding terms in the PFD are x−6
C
D
also linear, but is repeated twice, so the corresponding term in the PFD is x−1
+ (x+1)
2.
Ex+F
2
Finally, x + 3 is quadratic, so the corresponding term in the PFD is x2 +3 . Thus the
PFD is:
A
B
C
D
Ex + F
+
+
+
+ 2
.
2
x − 6 x + 1 x − 1 (x − 1)
x +3
MATH 152:561-563 – Spring 2015
Quiz #4 Solutions
Problem 2 (5 pts). Which substitution allows one to evaluate
(a) x − 6 =
√
2
R
√
dx
x2 −6x+5
?
5 sec θ
(b) x − 3 = 2 sec θ
(c) x − 6 = 2 sec θ
√
(d) x2 − 6x = 5 tan θ
(e) x − 3 = 2 tan θ
Solution. The correct answer is (b). First we complete the square under the radical and
write
(x2 − 6x) + 5 = (x2 − 6x + 9) + 5 − 9 = (x − 3)2 − 22 .
Z
dx
p
The integral then becomes
. This suggests that we make the substitu(x − 3)2 − 22
tion x − 3 = 2 sec θ.
MATH 152:561-563 – Spring 2015
Problem 3 (5 pts).
(a) 2 ln |x| +
(b)
2
x
2
x
R
x+2
dx
x2 (x−1)
Quiz #4 Solutions
3
=
+ 3 ln |x − 1| + C
+ 3 ln |x − 1| + C
(c) −3 ln |x| +
2
x
+ 3 ln |x − 1| + C
(d) − x2 + 3 ln |x − 1| + C
(e) −3 ln |x| −
2
x
+ 3 ln |x − 1| + C
Solution. The correct answer is (c). There are (at least) two ways to evaluate this integral:
integration by parts and integration by partial fractions. The method partial fractions is
slightly easier than integration by parts, so that is what we will do here. The degree of
the numerator is 1 and the degree of the denominator is 3, so we do not need to worry
about using long division. Note also that the denominator is factored completely, so we
can start to write the partial fraction decomposition (PFD). The factor x is linear and is
repeated twice, so the corresponding term in the PFD is Ax + xB2 (note that we could also
;
think of x2 as quadratic and then the corresponding term in the PFD would be Ax+B
x2
either way will give the same final answer). The term x − 1 is linear, so the corresponding
C
term in the PFD is x−1
. Thus the PFD is
C
x+2
A B
+ 2+
= 2
.
x x
x−1
x (x − 1)
This gives us the following equation:
x + 2 = Ax(x − 1) + B(x − 1) + Cx2 = (A + C)x2 (B − A)x − B.
Comparing coefficients on both sides, we see that A + C = 0, B − A = 1 and −B = 2.
From this we conclude that A = −3, B = −2 and C = 3. Thus, by the method of partial
fractions, the integral becomes
Z
Z x+1
3
2
3 =
−
−
+
dx
x2 (x − 1
x x2 x + 1
Z
Z
Z
1
1
1
= −3
dx − 2
dx + 3
dx
2
x
x
x+1
2
= −3 ln |x| + + 3 ln |x + 1| + C.
x
MATH 152:561-563 – Spring 2015
Quiz #4 Solutions
Problem 4 (5 pts). After an appropriate substitution, the integral
equivalent to which of the following?
R π/2
(a) 4 π/3 cos2 θdθ
(b) 4
R π/2
(c) 2
R π/2
π/6
π/6
(d) 2
(e) 4
R π/2
π/6
R2 √
√
3
4 − x2 dx is
sec θ tan2 θdθ
tan θdθ
R π/2
π/3
4
cos θdθ
cos2 θdθ
Solution.
The correct answer is (d). To see this, note that the radical is of the form
√
2
2
a − x which suggests either of the substitutions x = a sin θ (with − π2 ≤ θ ≤ π2 ) or
x = a cos θ (with 0 ≤ θ ≤ π). In this case, a = 2. If we make the substitution x = 2 sin θ,
then the integrand becomes
p
p
√
4 − 4 sin2 θ = 2 1 − sin2 θ = 2 cos2 θ = 2 cos θ.
√
√
√
When x = 3, we have 3 = 2 sin θ. That is, sin θ = 23 which means that θ =
√
sin−1 23 = π3 . When x = 2, then 2 = 2 sin θ. that is, sin θ = 1 which means that
θ = sin−1 1 = π2 . Thus, the integral becomes
Z
π/2
2
cos θdθ
π/3
which corresponds to choice (d). If we had made the substitution x = 2 cos θ, our final
integral would have been in terms of sin θ (with different limits of integration). Although
technically correct, this is not one of the answer choices, so this is not appropriate substitution.