SOLUTION 4: $ \ \ $ To integrate $ \displaystyle{ \int
{\sqrt{x^2+1} \over x } \ dx } $ use the trig substitution
$$ x = \tan \theta $$
so that
$$ dx = \sec^{2} \theta \ d \theta $$
Substitute into the original problem, replacing all
forms of x, getting
$$ \displaystyle { \int \frac{\sqrt{x^{2}+1}}{x} \ dx = \int
\frac{\sqrt{\tan^{2} \theta + 1}}{\tan \theta} \ \sec^{2} \theta \ d
\theta } $$
$$ = \displaystyle { \int \frac{\sqrt{\sec^{2} \theta}}{\tan
\theta} \sec^{2} \theta \ d \theta } $$
$$ = \displaystyle { \int \frac{\sec \theta}{\tan \theta} \
\sec^{2} \theta \ d \theta } $$
$$ = \displaystyle { \int \frac{\sec^{3} \theta}{\tan \theta} \ d
\theta } $$
$$ = \displaystyle { \int \frac{\sec \theta \sec^{2} \theta}{\tan
\theta} \ d \theta } $$
$$ = \displaystyle { \int \frac{\sec \theta (1 + \tan^{2}
\theta)}{\tan \theta} \ d \theta } $$
$$ = \displaystyle { \int \frac{ \cos \theta }{ \sin \theta } \cdot \frac{ 1 }{ \cos \theta }
\cdot (1 + \tan^{2} \theta ) \displaystyle \ d \theta } $$
$$ = \displaystyle { \int \frac{1 + \tan^{2} \theta}{\sin \theta} \
d \theta } $$
$$ = \displaystyle { \int \frac{ 1 }{ \sin \theta } \Big( 1 +
\Big( \frac{ \sin \theta }{ \cos \theta } \Big)^2 \Big) \ d \theta } $$
$$ = \displaystyle { \int \frac{ 1 }{ \sin \theta } \Big( 1 +
\frac{ \sin^2 \theta }{ \cos^2 \theta } \Big) \ d \theta } $$
$$ = \displaystyle { \int \Big(\frac{1}{\sin \theta} + \frac{\sin
\theta}{\cos^{2} \theta}\Big) \ d \theta } $$
$$ = \displaystyle { \int \Big(\csc \theta + \frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos
\theta} \Big) \ d \theta } $$
$$ = \displaystyle { \int (\csc \theta + \sec \theta \tan \theta ) \ d \theta } $$
(Recall that $ \displaystyle { \int \csc \theta \ d
\theta = \ln \Big|\csc \theta - \cot \theta \Big| + C } $ and $ \displaystyle { \int \sec \theta \tan \theta \ d
\theta = \sec \theta + C } $.)
$$ = \displaystyle { \ln \Big|\csc \theta - \cot \theta \Big| + \sec
\theta + C } $$
$\Big($We need to write our final answer in terms of $x$.
Since $ x = \tan \theta $ it follows that
$$ \tan \theta = \displaystyle{ x \over 1 } = \displaystyle{ opposite \over adjacent } $$
and from the Pythagorean Theorem that
$$ \displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2
\ \ \longrightarrow $$
$$ (1)^2 + (x)^2 = (hypotenuse)^2
\ \ \longrightarrow \ \ \ hypotenuse = \sqrt{x^2+1} \ \ \longrightarrow $$
$$ \sec \theta = \displaystyle{ hypotenuse \over adjacent }= \displaystyle{ \sqrt{x^2+1} \over 1 } \ , $$
$$ \csc \theta = \displaystyle{ hypotenuse \over opposite }= \displaystyle{ \sqrt{x^2+1} \over x } \ , $$
and
$$ \cot \theta = \displaystyle{ adjacent \over opposite }= \displaystyle{ 1 \over x } . \Big)$$
$$ = \displaystyle {\ln \Big|\frac{\sqrt{x^{2}+1}}{x} - \frac{1}{x} \Big|
+ \sqrt{x^{2}+1} + C } $$
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