tommywong

$\displaystyle \int_0^a \frac{arctan x}{1+ax} dx=\int_0^{arctan a} \frac{usec^2 udu}{1+atanu}=\frac{1}{\sqrt{1+a^2}}\int_0^{u_0} \frac{usec udu}{cos u_0 cos u+sin u_0 sin u}$

$\displaystyle =\frac{1}{\sqrt{1+a^2}} \int_0^{u_0} \frac{udu}{cos(u_0-u)cos u}$

$\displaystyle \int_0^{u_0} \frac{udu}{cos(u_0-u)cos u}=\int_{u_0}^0 \frac{(u_0-v)(-dv)}{cos(u_0-v)cos v}=\frac{u_0}{2}\int_0^{u_0} \frac{du}{cos(u_0-u)cos u}$

$\displaystyle \frac{1}{cos(u_0-u)cos u}=\frac{sin(u_0-u+u)}{sin u_0cos(u_0-u)cos u}=\frac{tan(u_0-u)+tan u}{sin u_0}$

$\displaystyle \int_0^{u_0} \frac{du}{cos(u_0-u)cos u}=\frac{-2ln cos u_0}{sin u_0}$

$\displaystyle \int_0^a \frac{arctan x}{1+ax} dx=\frac{1}{\sqrt{1+a^2}}\times\frac{u_0}{2}\times\frac{-2ln cos u_0}{sin u_0}$