CALCULUS 2 - QUIZ 6 KEY
RICKY NG
Give the form of the partial fraction decomposition. Do not solve for constants.
1.
x+1
.
(x2 +5x+6)(x2 +4)(x−3)2
Solution. Note that we can factor x2 + 5x + 6 = (x + 2)(x + 3), and x2 + 4 cannot be
factored any further. Thus
x+1
B
Cx + D
E
F
A
+
+ 2
+
+
=
.
2
2
2
(x + 5x + 6)(x + 4)(x − 3)
x+2 x+3
x +4
x − 3 (x − 3)2
2.
6
x2 (x2 −4)
Solution. Here x2 = x · x and x2 − 4 = (x − 2)(x + 2):
6
x2 (x2
− 4)
=
A B
C
D
+ 2+
+
.
x x
x−2 x+2
Find the partial fraction decomposition of the following.
3.
x+1
x3 −x2 −6x
Solution. First we decompose it and re-combine
x+1
A
B
C
= +
+
3
2
x − x − 6x
x x+2 x−3
A(x + 2)(x − 3) + Bx(x − 3) + Cx(x + 2)
=
.
x(x + 2)(x − 3)
This yields
A(x + 2)(x − 3) + Bx(x − 3) + Cx(x + 2) = x + 1.
It follows that
1
x = 0 ⇒ A(−6) = 1 ⇒ A = − ;
6
1
x = −2 ⇒ B(10) = −1 ⇒ B = − ;
10
4
x = 3 ⇒ C(15) = 4 ⇒ C = .
15
Therefore,
x+1
1 1
1
1
4
1
=
−
·
−
·
+
·
.
x3 − x2 − 6x
6 x 10 x + 2 15 x − 3
1
Integrate the following.
R
4. x1+x
2 +2x dx.
Solution. There are two ways to do it: u-sub and partial fraction decomposition.
For u-sub, let u = x2 + 2x, then du = 2x + 2 = 2(1 + x) dx. Thus
Z
Z
1+x
1
1
1
dx
=
du
=
ln |x2 + 2x| + C.
2
x + 2x
2
u
2
For partial fraction decomposition, note that
Z
Z
1+x
1 1 1
1
dx
=
·
+
·
dx
x2 + 2x
2 x 2 x+2
1
1
= ln |x| + ln |x + 2| + C.
2
2
5.
R
√1
x2 25−x2
dx.
√
Solution. This is a trig-sub with HYP = 5, OPS = x, and ADJ = 25 − x2 . So x = 5 sin θ
and dx = 5 cos θ dθ.Z
Z
1
1
√
dx =
· 5 cos θ dθ
(5 sin θ)2 5 cos θ
x2 25 − x2
Z
1
=
csc2 θ dθ
25
1
= − cot θ + C
25 √
1 25 − x2
+ C.
=−
25
x
R
Here you have to know csc2 θ dθ = − cot θ + C. Also, do not forget the proper
notation dθ.
2
Bonus:
R
ln(x2 − 1) dx.
Solution. There are two ways to solve it. Both involve integration by parts; one with
partial fraction decomposition; another one without but requires you to integrate by
parts twice. We have seen all of the techniques and ideas below in class and lab, but you
need to connect them in order to solve it.
Method 1: First integrate by parts: let u = ln(x2 − 1) and dv = dx, then du = x22x−1
and v = x. We get
Z
Z
2x2
2
2
dx.
ln(x − 1) dx = x ln(x − 1) −
x2 − 1
For the second integral, we need to apply partial fraction decomposition after we reduce
the degree of the numerator by simplifying or division:
2x2
2
1
1
=
2
+
−
.
=
2
+
2−1
x2 − 1
x
x
−
1
x
+
1
| {z }
p.f.d.
Finally, continue on the whole integral
Z
Z
2
2
ln(x − 1) dx = x ln(x − 1) −
2+
1
1
−
dx
x−1 x+1
= x ln(x2 − 1) − 2x − ln |x − 1| + ln |x + 1| + C.
Method 2: First use the property of the natural log function:
Z
Z
2
ln(x − 1) dx = ln(x + 1) + ln(x − 1) dx.
Apply integration by parts on each integral: Let u = ln(x + 1) and dv = dx, then
1
du = x+1
dx and v = x.
Z
Z
x
dx
ln(x + 1) dx = x ln(x + 1) −
x+1
Z
1
= x ln(x + 1) − 1 −
dx
x+1
= x ln(x + 1) − x + ln |x + 1| + C.
1
Likewise, let u = ln(x − 1) and dv = dx, then du = x−1
dx and v = x.
Z
Z
x
ln(x − 1) dx = x ln(x − 1) −
dx
x−1
Z
1
= x ln(x − 1) − 1 +
dx
x−1
= x ln(x − 1) − x − ln |x − 1| + C.
Hence,
Z
ln(x2 − 1) dx = x ln(x + 1) + x ln(x − 1) − 2x + ln |x + 1| − ln |x − 1| + C,
which is equivalent to the answer from Method 1.
3