Answers and Explanations 511
Answers
801–900
854.
116
15
Here’s the given integral:
x x
dx
+
∫
1
0
3
Begin by using the substitution u = x + 1 so that u – 1 = x and du = dx. Notice you can
find the new limits of integration by noting that if x = 3, then u = 3 + 1 = 4, and if x = 0,
then u = 0 + 1 = 1. With these new values, you have the following:
(
)
(
)
u
udu
u
u du
u
u
du
u
u
−
=
−
=
−
(
)
=
−
(
)
=
∫
∫
∫
1
1
2
5
2
3
2
1
4
1 2
1
4
3 2
1 2
1
4
5 2
3 2
1
4
5 5
4
2
3
4
2
5
1
2
3
1
64
5
16
3
2
5
2
3
116
15
5 2
3 2
5 2
3 2
−
(
) − −
(
)






=
−
− +
=
855.
2
1
7
1 2
4
7 3
4 3
( ) − +
− ( )
At first, you may have trouble seeing how a substitution will work for this problem.
But note that you can split x
5
into x
3
x
2
:
x x
dx
x x
x dx
5
3
3
0
1
3
3
3
2
0
1
1
1
+
=
+
∫
∫
Now you can use the substitution u = x
3
+ 1 so that u – 1 = x
3
and du = 3x
2 
dx, or
1
3
2
du x dx
=
. Notice that you can find the new limits of integration by noting that if
x = 1, then u = 1
3
+ 1 = 2, and if x = 0, then u = 0
3
+ 1 = 1. With these values, you
get the following integral:
1
3
1
1
3
1
3
3
7
3
4
1
3
3
7
2
1
2
1 3
4 3
1 3
1
2
7 3
4 3
1
2
7 3
(
)
u
u du
u
u
du
u
u
−
=
−
(
)
=
−
(
)
=
∫
∫
− −
(
) − −
(
)






=
−
− +
(
)
=
( ) −
3
4
2
3
7
1
3
4
1
1
3
3
7
2
3
4
2
3
7
3
4
2
4 3
7 3
4 3
7 3
4 3
7 3
1 1
7
1 2
4
4 3
+
− ( )