Integral of the Inverse Tangent

21 October 2013

Textbooks have a habit of throwing the most innocuous looking things at you, but it’s really quite an exercise to figure it out. In the section on Integration by Parts, particularly the section of exercises which only requires a single use of Integration by Parts.

∫tan⁡−1x dx\int\tan^{-1}x\,dx

Normally when integrating by parts, you deal with something in the form of f(x)⋅g(x)f(x)\cdot g(x), so to see something not in that form was a little unnerving. Except, then I realized it was in that form, because it’s really 1⋅tan⁡−1x1\cdot\tan^{-1}x. With this in mind, I gave it a go. I picked tan⁡−1x\tan^{-1}x for uu, largely because I can derive that using a formula.

∫tan⁡−1 dx=xtan⁡−1x−∫x1+x2 dxu=tan⁡−1xdv=dxdu=11+x2 dxv=x\begin{array}{cl} \int\tan^{-1}\,dx&=x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx\\ \begin{aligned} u&=\tan^{-1}x&dv&=dx\\ du&=\frac{1}{1+x^2}\,dx&v&=x \end{aligned} \end{array}

Now I’ll be perfectly frank, I am not a smart man, and I stared at x1+x2\frac{x}{1+x^2} far longer than I should have. What is the first thing you attempt when integrating? That lovely reversal of the chain rule, called u substitution. I actually had to call over Rob the Professor and he patiently waited for me to talk myself through the process until I saw it. If uu is set to be 1+x21+x^2, then everything works out smashingly.

∫x1+x2 dx=∫x2xu du=∫12⋅1u duu=1+x2du=2x dx⇒dx=12x du\begin{array}{cll} \int\frac{x}{1+x^2}\,dx&=\int\frac{x}{2xu}\,du&=\int\frac{1}{2}\cdot\frac{1}{u}\,du\\ \begin{aligned} u&=1+x^2\\ du&=2x\,dx\Rightarrow dx=\frac{1}{2x}\,du \end{aligned} \end{array}

Most will flinch at the middle step, but I felt it instructive to illustrate that the xx in the numerator cancels with the term in the denominator, which leaves the entire expression in terms of uu. I’ve kept the constant separate as well, because the integral of 1u\frac{1}{u} should be quite obvious, which will make the following step seem elementary.

∫12⋅1u du=12ln⁡∣u∣\int\frac{1}{2}\cdot\frac{1}{u}\,du=\frac{1}{2}\ln\left|u\right|

Finally, we reset the expression to be in terms of xx:

12ln⁡∣u∣=12ln⁡(1+x2)\frac{1}{2}\ln\left|u\right|=\frac{1}{2}\ln\left(1+x^2\right)

The absolute value is unnecessary because a positive constant plus the square of any number must be positive. When this expression is subtracted from the original leading term, we arrive at the solution to the integral of the inverse tangent:

∫tan⁡−1x dx=xtan⁡−1x−12ln⁡(1+x2)+C\int\tan^{-1}x\,dx=x\tan^{-1}x-\frac{1}{2}\ln\left(1+x^2\right)+C

This is a solution for §7.1 Exercise #19 from Calculus: Early Transcendentals.