Proving $\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$ 
How can I prove this?
$$\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$$

Here is my attempt
$$ \frac{1}{a^2b^2}\int_a^{b}\frac{\ln (1+x)}{(1+x)^2+(\frac{a+b}{ab})^2}\ dx$$
Applying the substitution $1+x=t \left (\dfrac{a+b}{ab}\right )$
$$ \frac{\frac{a+b}{ab}}{a^2b^2(\frac{a+b}{ab})^2}\int_{\frac{ab(a+1)}{a+b}}^{\frac{ab(b+1)}{a+b}}\frac{\ln (t)+\ln \left (\frac{a+b}{ab}\right )}{t^2+1}\ dt\\\\$$
$$ \frac{1}{ab(a+b)}\int_{\frac{ab(a+1)}{a+b}}^{\frac{ab(b+1)}{a+b}}\frac{\ln (t)+\ln \left (\frac{a+b}{ab}\right )}{t^2+1}\ dt\\\\$$
Again applying substitution $t\ =\dfrac{1}{u}$
$$ \frac{-1}{ab(a+b)}\int_{\frac{a+b}{ab(a+1)}}^{\frac{a+b}{ab(b+1)}}\frac{-\ln (u)+\ln \left (\frac{a+b}{ab}\right )}{u^2+1}\ du\\\\$$

By using integration by parts,
Differentiating ln(1+x) and integrating the remaining function gives an inverse tangent function integral in second part ,now set 1+x =z×(a+b)/ab  this substitution reduces the second integral to integration of arctan(z)/z under respective limits and observe that this integral is solvable by feynman parameter or integral under differential sign.
