Solving a Gateway question
Calculus II

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

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

We integrate by parts, setting

$$u := \ln (1+x^2), \qquad \frac{dv}{dx} := 1$$

and so

$$\frac{du}{dx} = \frac{2 x}{1+x^2}, \qquad v = x.$$

Therefore, we obtain

$$\begin{eqnarray*} I &=& x \ln (1+x^2) - 2 \int \frac{x^2}{1+x^2} dx \\ &=& x \l... ... \int \frac{1}{1+x^2} dx \\ &=& x \ln(1+x^2) - 2x + 2\arctan x. \end{eqnarray*}$$

We can check this:

$$\begin{eqnarray*} \frac{d}{dx} (x \ln(1+x^2) - 2x + 2\arctan x) &=& \ln(1+x^2) +... ...n(1+x^2) + \frac{2x^2 - 2(1+x^2) + 2}{1+x^2} \\ &=& \ln(1+x^2). \end{eqnarray*}$$

It is possible to obtain the same result using the substitution $x=\tan\theta$.