Answers and Explanations 509
Answers
801–900
Now substitute these values into the original integral:
sin( )
c os( )
cos
c os
u du
u
1
2
1
2
1
2
4
1
2
0
1
2
2
2
0
4
0
4
π
π
π
∫
= −
= −
( ) +
= −





 + 1 1
2
2 2
4
= − +
850.
3
1
2
1
1
2
tan
l n
−
+
+
(
) +
x
x
C
The given integral is
3
1
2
+
+
∫
x
x
dx
Begin by breaking the integral into two separate integrals:
3
1
1
3
1
1
2
2
2
2
+
+ +






=
+





 +
+






∫
∫
∫
x
x
x
dx
x
dx
x
x
dx
Notice that
3
1
3
2
1
+






=
+
∫
−
x
dx
x C
tan
.
All that’s left is to integrate the second integral; you can do this with a u-substitution.
For
x
x
dx
1
2
+






∫
, let u = 1 + x
2
so that du = 2x
 
dx, or 1
2
du x dx
=
. Substituting these
values into the second integral gives you the following:
1
2
1
1
2
1
2
1
2
u
du
u C
x
C
∫
=
+
=
+
+
ln
ln
The expression 1 + x
2
is always positive, so you can drop the absolute value sign:
=
+
(
) +
1
2
1
2
ln
x
C
Combining the solutions gives you
3
1
1
3
1
1
3
2
2
2
2
1
+
+ +






=
+





 +
+






=
∫
∫
∫
−
x
x
x
dx
x
dx
x
x
dx
x
tan
+ +
+
(
) +
1
2
1
2
ln
x
C