Solving a Gateway question
Calculus II

One question that came up on a Gateway exam was to find

$$I = \int \ln (1+x^2) dx.$$ (1)

To solve this, we need to exploit the following identities:

   

$$\sec^2 \theta 1 + \tan^2 \theta$$ = (2)

$$\frac{d \sec\theta}{d \theta} \sec \theta \tan \theta$$ = (3)

$$\frac{d \tan\theta}{d \theta} \sec^2 \theta.$$ = (4)

First substitute $x=\tan\theta$ in (1). Notice that $dx=\sec^2\theta d\theta$, from (4). We get

$$I = \int \sec^2\theta \ln (\sec^2\theta) d\theta,$$

using (2). We now integrate by parts, setting

$$u := \ln (\sec^2\theta), \qquad \frac{dv}{d\theta} := \sec^2 \theta$$

and so

$$\frac{du}{d\theta} = \frac{2 \sec\theta \sec\theta\tan\theta}{\sec^2\theta} = 2 \tan\theta, \qquad v = \tan\theta,$$

by the chain rule and from equations (3) and (4). Therefore, we obtain

$$\begin{eqnarray*} % latex2html id marker 36I &=& \ln (\sec^2\theta) \tan\theta... ...\mbox{from (\ref{dtan})} \\ &=& x \ln(1+x^2) - 2x + 2\arctan x. \end{eqnarray*}$$