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Quiz 3 — September 8, 2010 – Section 10 – 11:15 – 12:05
Remove everything from your desk except this page and a pencil or pen.
Circle
your answer.
Show your work. Check
your answer.
The quiz is worth 5 points.
Find
i
√
x
2
+ 2
xdx
.
Answer:
We complete the square:
x
2
+ 2
x
= (
x
2
+ 2
x
+ 1)

1 = (
x
+ 1)
2

1.
Let
x
+ 1 = sec
θ
. It follows that (
x
+ 1)
2

1 = tan
2
θ
and
dx
= sec
θ
tan
θdθ
. The
original problem is equal to
I
tan
2
θ
sec
θdθ.
We use integration by parts. Let
u
= tan
θ
and
dv
= sec
θ
tan
θdθ
. It follows that
du
= sec
2
θdθ
and
v
= sec
θ
. So
I
tan
2
θ
sec
θdθ
= sec
θ
tan
θ

I
sec
3
θdθ
= sec
θ
tan
θ

I
(tan
2
θ
+ 1) sec
θdθ
= sec
θ
tan
θ

I
sec
θdθ

I
tan
2
θ
sec
θdθ.
Add
i
tan
2
θ
sec
θdθ
to both sides to see that
2
I
tan
2
θ
sec
θdθ
= sec
θ
tan
θ

I
sec
θdθ.
So
I
r
x
2
+ 2
xdx
=
I
tan
2
θ
sec
θdθ
= (1
/
2)
b
sec
θ
tan
θ

I
sec
θdθ
B
= (1
/
2) [sec
θ
tan
θ

ln

sec
θ
+ tan
θ

] +
C
=
(1
/
2)
±
(
x
+ 1)
r
x
2
+ 2
x

ln

(
x
+ 1) +
r
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 Fall '11
 KUSTIN
 Calculus, Equals sign, tan2 θ sec

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