SOLUTION 11: $ \ \ $ To integrate $ \displaystyle{ \int { x
\over \sqrt{x^4-16} } \ dx = \int { x \over \sqrt{(x^2)^2-16} } \ dx } $ begin with the ordinary u-substitution
$$ u = x ^{2} $$
so that
$$ du = 2x \ dx $$
Substitute into the original problem, replacing all
forms of x, getting
$$ \displaystyle { \int \frac{x}{\sqrt{x^{4}-16}} \ dx = \int
\frac{1/2}{\sqrt{u^{2}-16}} \ du } $$
Now use the trig substitution
$$ u = 4 \sec \theta $$
so that
$$ du = 4 \sec \theta \tan \theta \ d \theta $$
Substitute into the original problem, replacing all
forms of u, getting
$$ = \displaystyle { \int
\frac{1/2}{\sqrt{(4 \sec \theta )^{2}-16}} \ du } $$
$$ = \displaystyle { \int \frac{1/2}{\sqrt{16 \sec^{2} \theta -
16}} \cdot 4 \sec \theta \tan \theta \ d \theta } $$
$$ = \displaystyle { \int \frac{1/2}{\sqrt{16 ( \sec^{2} \theta -
1)}} \cdot 4 \sec \theta \tan \theta \ d \theta } $$
$$ = \displaystyle { 2 \int \frac{\sec \theta \tan \theta}{\sqrt{16
\tan^{2} \theta}} \ d \theta } $$
$$ = \displaystyle { 2 \int \frac{\sec \theta \tan \theta}{4 \tan
\theta} \ d \theta } $$
$$ = \displaystyle { \frac{1}{2} \int \sec \theta \ d \theta } $$
$$ = \displaystyle { \frac{1}{2} (\ln \Big|\sec \theta + \tan \theta \Big| +
C) } $$
$ \Big( $ We need to write our final answer in terms of $x$.
Since $ u = 4 \sec \theta $ it follows that
$$ \sec \theta = \displaystyle{ u \over 4 } = \displaystyle{ hypotenuse \over adjacent } $$
and from the Pythagorean Theorem that
$$ \displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2
\ \ \longrightarrow $$
$$ (4)^2 + (opposite)^2 = (u)^2
\ \ \longrightarrow \ \ \ opposite = \sqrt{u^2-16} \ \ \longrightarrow $$
$$ \tan \theta = \displaystyle{ opposite \over adjacent }= \displaystyle{ \sqrt{u^2-16} \over 4 } . \Big) $$
$$ = \displaystyle { \frac{1}{2} (\ln \Big|{ u \over 4 } + { \sqrt{u^2-16} \over 4 }\Big| +
C) } $$
(Now use the fact that $ u=x^2 $.)
$$ = \displaystyle { \frac{1}{2} (\ln \Big|\frac{x^{2}}{4} +
\frac{\sqrt{x^{4}-16}}{4} \Big| + C) } $$
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