<s>[INST] 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 $t\ =\dfrac{1}{u}$
$$ \frac{-1}{ab(a+b)}\int_{\frac{a+b}{ab(a+1)}}^{\frac{a+b}{ab(b+1)}}\frac{-\ln (u)+\ln \left (\frac{a+b}{ab}\right )}{u^2+1}\ du\\\\$$
 [/INST] Repeating your steps for the antiderivarive ( assuming $a>0$ and $b >0$)$$\int\frac{\log (1+x)}{a^2b^2(1+x)^2+({a+b})}\, dx=\frac 1{a^2\,b^2}\int\frac{\log (1+x)}{(1+x)^2+k^2}\, dx$$ with $k=\sqrt{\frac {a+b}{a^2\,b^2}}$ Now, let $r$ and $s$ be the complex roots of the denominator and use
$$\frac{1}{(1+x)^2+k^2}=\frac 1{(x-r)(x-s)}=\frac 1{r-s}\Big[\frac{1}{x-r}-\frac{1}{x-s} \Big]$$ So, we face two integrals
$$I_t=\int \frac{\log(1+x)}{x-t}\,dx=\text{Li}_2\left(\frac{x+1}{t+1}\right)+\log (x+1) \log
   \left(\frac{t-x}{t+1}\right)$$ Thsi would lead to much complex formulae for the definite integral (in which will remain logarithms, polylogarithms (with complex arguments) and  arctangents). If I am not mistaken, $\log(a+b)$ would never appear.
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