Part II: The Answers
512
Answers
801–900
856.
6
11
3
18
5
3
11 6
5 6
(
)
(
)
x
x
C
+
−
+
+
Here’s the given integral:
x
x
dx
+
∫
3
6
If you let u = x + 3, then du = dx and u – 3 = x. Substituting into the original integral, you
get the following:
(
)
(
)
u
u
du
u
u
du
u
u
u
du
u
u
du
u
−
=
−
=
−






=
−
(
)
=
∫
∫
∫
∫
−
3
3
3
3
6
1 6
1 6
1 6
5 6
1 6
1 11 6
5 6
11 6
5 6
11 6
3 5 6
6
11
3
18
5
3
−








+
=
+
−
+
+
u
C
x
x
C
(
)
(
)
857.
1
10
1
1
6
1
4
5 2
4
3 2
x
x
C
+
(
) −
+
(
) +
At first glance, substitution may not seem to work for this problem. However, you can
rewrite the integral as follows:
x x
dx
x x x
dx
7
4
4 3
4
1
1
+
=
+
∫
∫
If you let u = x
4
+ 1, then you get du = 4x
3 
dx, or 1
4
3
du x dx
=
. Notice also from the
u- substitution that u – 1 = x
4
. Substitute into the original integral and simplify:
(
)
u
u
du
u
u
du
u
u
−
( ) =
−
(
)
=








−








∫
∫
1
1
4
1
4
1
4
5 2
3 2
3 2
1 2
5 2
3 2
 







+
=
+
(
) −
+
(
) )
(
+
=
+
(
) −
+
C
x
x
C
x
x
1
4
2
5
1
2
3
1
1
10
1
1
6
1
4
5 2
4
3 2
4
5 2
4
( (
) +
3 2
C
858.
1
4
4
1
16
4
x
x
x C
sin( )
cos( )
+
+
The given integral is
x
x dx
cos( )
4
∫
Use integration by parts with u = x so that du = dx and let dv = cos(4x)
 
dx so that
v
x
=
( )
1
4
4
sin
. Using the integration by parts formula gives you the following: