Answers and Explanations 487
Answers
701–800
794.
ln
tanh
2 +
+
x C
The given problem is
sech
tanh
2
2
x
x
dx
+
∫
Begin by using the substitution u = 2 + tanh x so that du = sech
2 
x
 
dx:
1
2
u
du
u C
x C
∫
=
+
=
+
+
ln
ln
tanh
795.
1
6
6
1
36
6
x
x
x C
sinh
cosh
( )−
( )+
Here’s the given problem:
x
x dx
cosh 6
( )
∫
To find the antiderivative, use integration by parts with u = x so that du = dx, and let
dv = cosh(6x)
 
dx so that v
x
=
( )
1
6
6
sinh
. This gives you the following:
x
x dx
x
x
x dx
x
x
cosh
sinh
sinh
sinh
co
6
1
6
6
1
6
6
1
6
6
1
36
( )
=
( )
(
) −
( )
=
( )−
∫
∫
s sh 6x C
( )+
796.
−





 +
5
2
5
2
coth x
C
Here’s the given problem:
x
x dx
csch
2
2
5






∫
To find the antiderivative, begin by using the substitution u x
=
2
5
so that du
x dx
= 2
5
,
or 5
2
du x dx
=
. With these substitutions, you get the following:
5
2
5
2
5
2
5
2
2
csch
coth
coth
u du
u
C
x
C
( ) = −
( )
(
) +
= −





 +
∫
797.
1
2
1
2
csch x
C





 +
Start with the given problem:
csch
coth
1
1
2
2
3
x
x
x
dx












∫
To find the antiderivative, begin by using the substitution u x
= 1
2
so that du = –2x
−3
,