Part II: The Answers
480
Answers
701–800
773.
1
2
4 2
2
2
1
ln
tan
x
x C
+ +
( ) +
−
Begin by splitting the integral:
x
x
dx
x
x
dx
x
dx
+
+
=
+
+
+
∫
∫
∫
4
4
4
4
4
2
2
2
For the first integral, you can use the substitution u = x
2
+ 4 so that du = 2x
 
dx, or
1
2
du x dx
=
. For the second integral, simply use the tan
−1
x formula for integration,
which gives you the following:
1
2
1
4
1
2
1
2
4 1
2
2
1
2
4 2
2
2
2
1
2
1
u
du
x
dx
u
x C
x
x
+
+
=
+ (
) +
=
+ +
(
∫
∫
−
−
ln
tan
ln
tan
) ) + C
774.
4 2 3 6
−
+ π
Here’s the given problem:
2
3
4
2
2
3
x
x x
dx
−
−
∫
Begin by completing the square on the expression under the radical:
4
4
4
4
4 4
2
4
4
2
2
2
2
2
2
2
x x
x
x
x
x
x
x
x
x
−
= − +
= −
−
(
)
= −
−
+
(
) +
= − −
(
) +
= − −
(
)
This gives you
2
3
4
2
3
4
2
2
2
3
2
2
3
x
x x
dx
x
x
dx
−
−
=
−
− −
∫
∫
(
)
Now use the substitution u = x – 2 so that u + 2 = x and du = dx. Notice you can find
the new limits of integration by noting that if x = 3, then u = 3 – 2 = 1, and if x = 2, then
u = 2 – 2 = 0.
2
3
4
2
2
2 3
4
2
2
3
2
0
1
x
x
dx
u
u
du
−
− −
=
+
(
)−
−
∫
∫
(
)
Next, use some algebra to manipulate the expression under the radical: