Part II: The Answers
570
Answers
901–1,001
959.
ln x
x
C
+ −
+ +
+
1 1
1 1
Here’s the given problem:
1
1
x x
dx
+
∫
Begin by using the rationalizing substitution u
x
=
+1 so that u
2
= x + 1, or u
2
– 1 = x,
and 2u
 
du = dx. This substitution gives you
2
1
2
1
1
1
2
u du
u
u
u
u
du
−
(
)
=
−
+
∫
∫ ( )( )
Next, find the fraction decomposition:
1
1
1
1
1
(
)(
)
u
u
A
u
B
u
−
+
= −
+ +
Multiply both sides by (u – 1)(u + 1):
1
1
1
=
+ +
−
A u
B u
(
)
(
)
Notice that if u = –1, you get 1 = B(–2) so that − =
1
2
B , and if u = 1, you get 1 = A(2) so
that 1
2
= A. With these values, you get the following:
2
1
1
1
2
1 2
1
1 2
1
1
1
(
)(
)
ln
ln
ln
u
u
du
u
u
du
u
u
C
u
−
+
=
−
− +








=
− −
+ +
=
−
∫
∫
1 1
1
1 1
1 1
u
C
x
x
C
+
+
=
+ −
+ +
+
ln
960.
2
3
2
+ ln
Here’s the given expression:
x
x
dx
−
∫ 1
4
9
Begin by using the rationalizing substitution u
x
=
so that u
2
= x and 2u
 
du = dx.
Note that if x = 9, you have u =
=
9 3, and if x = 4, you have u =
=
4 2, producing the
following integral:
u u
u
du
u
u
du
du
u
u
du
( )
( )
(
)(
)
2
1
2
1 1
1
2 1
2
1
1
1
2
2
3
2
2
2
3
2
3
2
3
−
=
− +
−
=
+
−
+
∫
∫
∫
∫ ∫
Next, perform the fraction decomposition:
1
1
1
1
1
(
)(
)
u
u
A
u
B
u
−
+
= −
+ +