Part II: The Answers
564
Answers
901–1,001
For the first term in the integrand, simply use an elementary antiderivative:
3
1
3
1
x
dx
x
C
+
=
+ +
∫
ln
. For the second term in the integrand, use a substitution on
3
1
2
(
)
x
dx
+
∫
where u = x + 1 so that du = dx. Using these values gives you
2
2
2 1
2
2
1
2
2
1
u
du
u du
u
C
u
C
x
C
∫
∫
=
= −
+
= − +
= − +
+
−
−
Therefore, the answer is
3
1
2
1
3
1
2
1
2
x
x
dx
x
x
C
+
+
+






=
+ − +
+
∫
(
)
ln
Note: You can also evaluate
3
5
1
2
x
x
dx
+
+
∫ ( )
with the substitution u = x + 1 so that du = dx
and u – 1 = x.
955.
x
x
x
x
C
2
2
14
5
3 9
5
2
+ +
− −
+ +
ln
ln
The given expression is
x
x
x
x
dx
3
2
6
5
6
−
+
− −
∫
The degree of the numerator is greater than or equal to the degree of the denominator,
so use polynomial long division to get the following:
x
x
x
x
dx
x
x
x
x
dx
3
2
2
6
5
6
1
11
6
−
+
− −
=
+ +
+
− −






∫
∫
Next, perform a fraction decomposition:
x
x
x
A
x
B
x
+
−
+
= −
+ +
11
3
2
3
2
(
)(
)
Multiply both sides of the equation by (x – 3)(x + 2):
x
A x
B x
+ =
+ +
−
11
2
3
(
)
(
)
If you let x = 2, you get 9 = B(–2 – 3) so that − =
9
5
B. And if you let x = 3, you get
14 = A(3 + 2) so that 14
5
= A. With these values, you produce the following integral:
x
x
x x
dx
x
x
x
dx
+ +
+
− −






=
+ + −
+
−
+








∫
∫
1
11
6
1
14 5
3
9 5
2
2
To integrate, use basic antiderivative formulas to get
x
x
x
dx x
x
x
x
C
+ + −
+
−
+








=
+ +
− −
+ +
∫
1
14 5
3
9 5
2
2
14
5
3 9
5
2
2
ln
ln