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 $$\cot \theta = \displaystyle{ adjacent \over opposite }= \displaystyle{ 1 \over x } . \Big)$$ $$= \displaystyle {\ln \Big|\frac{\sqrt{x^{2}+1}}{x} - \frac{1}{x} \Big| + \sqrt{x^{2}+1} + C }$$