Evaluating $\int_0^2(\tan^{-1}(\pi x)-\tan^{-1} x)\,\mathrm{d}x$

Use the change of variables $y = xt.$

$$I = \int_0^2 \int_x^{\pi x} \frac { \mathrm{d}y \, \mathrm{d}x} {y^2+1}\\=\int_0^2 \int_1^{\pi } \frac { x} {x^2t^2+1}\mathrm{d}t \, \mathrm{d}x\\=\int_1^{\pi} \int_0^{2 } \frac { x} {x^2t^2+1}\mathrm{d}x \, \mathrm{d}t\\=\int_1^{\pi} \frac{\ln(1+4t^2)}{2t^2} \mathrm{d}t$$

Now use integration by parts.

$$I = -\left.\frac{\ln(1+4t^2)}{2t}\right|_1^{\pi}+4\int_1^{\pi}\frac1{1+4t^2} \, dt$$

A more straight forward approach uses integration by parts.

Define: \begin{align} & I(c)=\int_{a}^{b}dx(1 \times \arctan{c x})=\int_{ac}^{bc}\frac{dy}{c} (1 \times\arctan{ y})=\\&\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\int_{ac}^{bc}\frac{y}{1+y^2} \end{align}

using partial fraction this reads: \begin{align} I(c)=\frac{1}{c}y \arctan(y)|_{ac}^{bc}-\frac{1}{2c}\log(1+y^2)|_{ac}^{bc} \end{align}

taking $I(\pi)-I(0)$ with $a=0$ and $b=2$ we are done