Answers and Explanations 553
Answers
901–1,001
First complete the square on the quadratic expression:
x
x
x
x
x
x
x
2
2
2
2
4
2
4
2
4
4 2 4
2
2
+
+ =
+
(
) +
=
+
+
(
) + −
=
+
−
(
)
So the integral becomes
dx
x
x
dx
x
2
2
4
2
2
2
+
+
(
)
=
+
−
∫
∫ (
)
Use the substitution (
)
sec
x + =
2
2
θ , which gives you dx
d
= 2 sec tan
θ
θ θ . Putting
this value into your integral yields
dx
x
d
d
(
)
sec tan
sec
sec tan
sec
+
−
(
)
= (
) −
(
)
=
−
∫
∫
2
2
2
2
2
2
2
2
2
2
2
θ
θ θ
θ
θ
θ θ
θ
( (
)
=
−
(
)
(
)
=
(
)
(
)
=
∫
∫
∫
2
2
1
2
2
2
2
2
2
sec tan
sec
sec tan
tan
sec
θ
θ θ
θ
θ
θ θ
θ
d
d
θ θ θ
θ
d
tan
∫
Because sec
tan
cos
cos
sin
csc
θ
θ
θ
θ
θ
θ
=
=
1
, you have
2
2
2
2
2
2
sec
tan
csc
ln csc
cot
θ θ
θ
θ θ
θ
θ
d
d
C
∫
∫
=
=
−
+
From the substitution x +
(
)=
2
2secθ , or
(
) sec
x + =
2
2
θ , you can deduce that
cscθ =
+
+
+
x
x
x
2
4
2
2
and that cotθ =
+
+
2
4
2
2
x
x
. (To deduce these values for cscθ
and cotθ , you can label a right triangle using
x +
(
) =
2
2
secθ and then find the missing
side using the Pythagorean theorem.) Putting these values into the antiderivative gives
you the solution:
2
2
2
4
2
2
4
2
2
2
2
2
4
2
2
2
2
ln
ln
x
x
x
x
x
C
x
x
x
C
+
+
+
−
+
+
+
=
+ −
+
+
+