Answer
$$ = \frac{{5{x^2} - 2x + 3}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\ln {{\left( {x - 1} \right)}^3}{{\left( {{x^2} + 1} \right)}^4}} \right] \cr
& {\text{logarithm of a product}} \cr
& \frac{d}{{dx}}\left[ {\ln {{\left( {x - 1} \right)}^3}} \right] + \left[ {\ln {{\left( {{x^2} + 1} \right)}^4}} \right] \cr
& {\text{power rule for logarithms}} \cr
& \frac{d}{{dx}}\left[ {3\ln \left( {x - 1} \right)} \right] + \left[ {4\ln \left( {{x^2} + 1} \right)} \right] \cr
& 3\frac{d}{{dx}}\left[ {\ln \left( {x - 1} \right)} \right] + \left[ {\ln \left( {{x^2} + 1} \right)} \right] \cr
& {\text{differentiate}} \cr
& = 3\left( {\frac{1}{{x - 1}}} \right) + \frac{{2x}}{{{x^2} + 1}} \cr
& = \frac{3}{{x - 1}} + \frac{{2x}}{{{x^2} + 1}} \cr
& = \frac{{3{x^2} + 3 + 2{x^2} - 2x}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} \cr
& = \frac{{5{x^2} - 2x + 3}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} \cr} $$
