Part II: The Answers
478
Answers
701–800
Use the substitution u = ln x so that du x
dx
= 1 :
1
1
1
1
2
2
1
1
x
x
dx
u
du
u C
x C
+ ( )
(
)
= +
=
+
=
( )+
∫
∫
−
−
ln
tan
tan ln
769.
2
1
tan
−
+
x C
Here’s the given problem:
dx
x
x
1 +
(
)
∫
Begin with the substitution u
x
=
so that u
2
= x and 2u
 
du = dx. Using these substitutions gives you the following:
dx
x
x
u
u
u
du
u
du
u C
x C
1
2
1
2
1
1
2
2
2
2
1
1
+
(
)
=
+
(
)
=
+
=
+
=
+
∫
∫
∫
−
−
tan
tan
770.
1
5
5
1
sec
−
+
x C
Here’s the given problem:
1
25
2
x x
dx
−
∫
Begin by doing a bit of algebra to manipulate the denominator of the integrand:
1
25
1
25 25
1
1
5
5
1
2
2
2
x x
dx
x
x
dx
x x
dx
−
=
−






=
( ) −
∫
∫
∫
Now use the substitution u x
= 5
so that du
dx
= 1
5
, or 5du = dx:
1
5
5
1
5
5 5
1
1
5
1
1
1
5
1
5
2
2
2
1
1
x x
dx
du
u u
u u
du
u C
( ) −
=
( ) −
=
−
=
+
=
∫
∫
∫
−
−
sec
sec x x C
5
+