Calculus BC Trigonometric Substitition

[hwmaltby]

1. The problem statement, all variables and given/known data

\int \frac {\sqrt {1+x^2}}{x} \, dx

2. Relevant equations

\sin^2{\theta}+\cos^2{\theta}=1

3. The attempt at a solution

x=\tan{\theta}

I simplified it down to:

\int \frac {1}{\sin{\theta} \cdot \cos^2{\theta}} \, d\theta

Which I do not know how to integrate.

Any help would be wonderful!

[hwmaltby]

Never mind... I rewrote it as:

\int \csc{\theta}\cdot\sec^2{\theta} \, d\theta

And used u-substition:

u=\sec{\theta}

\int \frac{\csc{\theta} \cdot \sec^2{\theta}}{\tan{\theta} \cdot \sec{\theta}} \, du

\int \csc^2{\theta} \, du

\int \frac{u^2}{u^2-1} \, du

\frac{1}{2} \cdot \ln |\frac{u-1}{u+1}| +u+C

I then simplified using my previous substitutions and logarithm rules to get my final answer:

\ln |\frac{\sqrt{1+x^2}-1}{x}|+\sqrt{1+x^2}+C

[Mark44]

And you can check by differentiating your answer. If you end up with sqrt(1 + x^2)/x, you're golden.