Find $\int_0^2 \arctan(\pi x)-\arctan(x)\, \mathrm dx$ Find $$\int_0^2 \arctan(\pi x)-\arctan(x)\, \mathrm dx$$
The hint is also given : Re-write the Integrand as an Integral
I think we have to Re-write  this single integral as a double integral and then do something.
Any one any ideas 
 A: I think the hint is
$$\arctan\pi x -\arctan x=\int_x^{\pi x}\frac{du}{1+u^2}$$
Be careful with how you handle the variables.
A: We'll generalize it a bit$$I(\alpha )=\int_0^2 \arctan(\alpha  x)\, \mathrm ddx$$
And we have $I(0)=0$
$$I'(\alpha )=\int_0^2 \frac{x}{1+\alpha ^2x^2}\, \mathrm dx=\left[\frac{1}{2\alpha ^2}\ln (1+\alpha ^2x^2)\right]_0^2=\frac{\ln(1+4\alpha ^2)}{2\alpha ^2}$$
$$\begin{align}
I(\alpha )&=2\arctan(2\alpha )-\frac{\ln(4\alpha ^2+1)}{2\alpha }+c\\
&=2\arctan(2\alpha )-\frac{\ln(4\alpha ^2+1)}{2\alpha }\\
\end{align}$$

$$\large I(\alpha )=2\arctan(2\alpha )-\frac{\ln(4\alpha ^2+1)}{2\alpha }$$

And Finally use
$$\int_0^2\arctan(\pi x)-\arctan( x)\, \mathrm dx=I(\pi)-I(1)$$
To get
$$\begin{align}
\int_0^2\arctan(\pi x)-\arctan( x)\, \mathrm dx&=\frac{1}{2}\left[4\arctan(2\pi )-\frac{\ln(4\pi^2+1)}{\pi}-4\arctan(2)+\ln(5)\right]\\
&\approx 0.8274\\
\end{align}$$

A: $\displaystyle \arctan\pi x -\arctan x=\int_x^{\pi x}\frac{du}{1+u^2}=\int_1^{\pi} \dfrac{x}{1+x^2u^2}du$
Therefore:
$I=\displaystyle \int_0^2 \arctan(\pi x)-\arctan(x)\, \mathrm{ dx}=\int_0^2\int_1^{\pi} \dfrac{x}{1+x^2u^2}\mathrm{ dx}\mathrm{ du}=\int_1^{\pi}\Big(\int_0^2\dfrac{x}{1+x^2u^2}\mathrm{ dx}\Big)\mathrm{ du}$
$\displaystyle \int_0^2\dfrac{x}{1+x^2u^2}\mathrm{ dx}=\dfrac{1}{2u^2}\Big[\ln(1+x^2u^2)\Big]_0^2=\dfrac{\ln(1+4u^2)}{2u^2}$
Therefore:
$I=\displaystyle \int_1^{\pi} \dfrac{\ln(1+4u^2)}{2u^2}\mathrm{ du}=\Big[-\dfrac{1}{2u}\ln(1+4u^2)\Big]_1^{\pi}+\int_1^{\pi}\dfrac{8u}{1+4u^2}\times\dfrac{1}{2u}\mathrm{ du}$
$I=\displaystyle \dfrac{1}{2}\ln 5-\dfrac{1}{2\pi}\ln(1+4\pi^2)+\Big[2\arctan (2u)\Big]_1^{\pi}$
$I=\dfrac{1}{2}\ln 5-\dfrac{1}{2\pi}\ln(1+4\pi^2)+2\arctan(2\pi)-2\arctan 2$
