Answer
$$ - 3\frac{{3x - 1}}{{9{x^2} + 1}} + 2x\frac{{{x^2} - 1}}{{{x^4} + 1}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\int_{3x}^{{x^2}} {\frac{{t - 1}}{{{t^2} + 1}}} dt} \right] \cr
& {\text{Rewriting}} \cr
& \frac{d}{{dx}}\left[ {\int_{3x}^{{x^2}} {\frac{{t - 1}}{{{t^2} + 1}}} dt} \right] = \frac{d}{{dx}}\left[ {\int_{3x}^0 {\frac{{t - 1}}{{{t^2} + 1}}} dt} \right] + \frac{d}{{dx}}\left[ {\int_0^{{x^2}} {\frac{{t - 1}}{{{t^2} + 1}}} dt} \right] \cr
& {\text{Apply }}\frac{d}{{dx}}\left[ {\int_{g\left( x \right)}^a {f\left( t \right)dt} } \right] = - f\left( {g\left( x \right)} \right)g'\left( x \right) \cr
& \frac{d}{{dx}}\left[ {\int_{3x}^{{x^2}} {\frac{{t - 1}}{{{t^2} + 1}}} dt} \right] = - \frac{{\left( {3x} \right) - 1}}{{{{\left( {3x} \right)}^2} + 1}}\left( 3 \right) + \frac{{{x^2} - 1}}{{{{\left( {{x^2}} \right)}^2} + 1}}\left( {2x} \right) \cr
& {\text{Simplifying}} \cr
& \frac{d}{{dx}}\left[ {\int_{3x}^{{x^2}} {\frac{{t - 1}}{{{t^2} + 1}}} dt} \right] = - 3\frac{{3x - 1}}{{9{x^2} + 1}} + 2x\frac{{{x^2} - 1}}{{{x^4} + 1}} \cr} $$
