Answers and Explanations 479
Answers
701–800
771.
sec
−
− +
1
1
x
C
Here’s the given problem:
1
1
2
2
x
x
x
dx
−
(
)
−
∫
Begin by completing the square on the expression underneath the square root.
x
x
x
x
x
2
2
2
2
2 1 1
1
1
−
=
−
+
(
) −
=
−
(
) −
This gives you
1
1
2
1
1
1
1
2
2
x
x
x
dx
x
x
dx
−
(
)
−
=
−
(
)
−
(
) −
∫
∫
Now use the substitution u = x – 1 so that du = dx:
1
1
1
1
1
1
2
2
1
1
x
x
dx
du
u u
u C
x
C
−
(
)
−
−
=
−
=
+
=
− +
∫
∫
−
−
(
)
sec
sec
772.
1
3
1
3
sin
C
−
( ) +
e
x
The given problem is
e
e
dx
x
x
3
6
1 −
∫
You’d like to have “1 – (a quantity squared)” under the radical so that you can use the
formula
1
1
2
1
−
=
+
−
∫
x
dx
x C
sin
, so you may want to write the second term under the
radical as a quantity squared and try a substitution to see whether it all works out.
(Of course, you can use the substitution without rewriting, but rewriting may help
you see which substitution to use.)
e
e
dx
e
e
dx
x
x
x
x
3
6
3
3
2
1
1
−
=
− ( )
∫
∫
Now you can use the substitution u = e
3x
so that du = 3e
3x 
dx, or 1
3
3
du e dx
x
=
:
e
e
dx
u
du
u C
e
x
x
x
3
3
2
2
1
1
3
1
1
3
1
1
1
3
1
3
− ( )
=
−
=
+
=
+
∫
∫
−
−
sin
sin (
) C