Integral of 1/(a^2+x^2)

The term 1/(a² + x²) is purely algebraic. However, its integral is trigonometric. When you look this up in a table of integrals, you'll find that:

\int \frac{1}{a^2+x^2}\mathrm{d}x=\frac{1}{a}\arctan \left ( \frac{x}{a} \right )+C

How is this so?

This is where Pythagora's Theorem and the rules of trigonometric come in to help.

We can't simply use a u-substitution to solve this problem. Instead, if we construct a right-angle triangle with the 2 shorter sides of lengths a and x, then the relationship between the hypotenuse and the the 2 shorter sides is a²+x² (by Pythagoras' Theorem).

Now...

\tan\theta =\frac{x}{a}

Or...

x=a\tan \theta

Taking the derivative with respect to θ, we get...

\frac{\mathrm{d} x}{\mathrm{d} \theta }=\sec ^2\theta

Thus we need to substitute x = atanθ, and dx = sec²θdθ to solve the integral.