Answer
$$\eqalign{
& \left( {\text{a}} \right)3{x^2}{\sin ^2}\left( {{x^3}} \right) - 2x{\sin ^2}\left( {{x^2}} \right) \cr
& \left( {\text{b}} \right)\frac{2}{{1 - {x^2}}} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\frac{d}{{dx}}\int_{{x^2}}^{{x^3}} {{{\sin }^2}t} dt \cr
& {\text{Apply }}\frac{d}{{dx}}\int_{h\left( x \right)}^{g\left( x \right)} {f\left( t \right)dt} = f\left( {g\left( x \right)} \right)g'\left( x \right) - f\left( {h\left( x \right)} \right)h'\left( x \right) \cr
& \frac{d}{{dx}}\int_{{x^2}}^{{x^3}} {{{\sin }^2}t} dt = {\sin ^2}\left( {{x^3}} \right)\frac{d}{{dx}}\left[ {{x^3}} \right] - {\sin ^2}\left( {{x^2}} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{Differentiate and simplify}} \cr
& \frac{d}{{dx}}\int_{{x^2}}^{{x^3}} {{{\sin }^2}t} dt = {\sin ^2}\left( {{x^3}} \right)\left( {3{x^2}} \right) - {\sin ^2}\left( {{x^2}} \right)\left( {2x} \right) \cr
& \frac{d}{{dx}}\int_{{x^2}}^{{x^3}} {{{\sin }^2}t} dt = 3{x^2}{\sin ^2}\left( {{x^3}} \right) - 2x{\sin ^2}\left( {{x^2}} \right) \cr
& \cr
& \left( {\text{b}} \right)\frac{d}{{dx}}\int_{ - x}^x {\frac{1}{{1 + t}}} dt \cr
& {\text{Apply }}\frac{d}{{dx}}\int_{h\left( x \right)}^{g\left( x \right)} {f\left( t \right)dt} = f\left( {g\left( x \right)} \right)g'\left( x \right) - f\left( {h\left( x \right)} \right)h'\left( x \right) \cr
& \frac{d}{{dx}}\int_{ - x}^x {\frac{1}{{1 + t}}} dt = \frac{1}{{1 + x}}\frac{d}{{dx}}\left[ x \right] - \frac{1}{{1 - x}}\frac{d}{{dx}}\left[ { - x} \right] \cr
& {\text{Differentiate and simplify}} \cr
& \frac{d}{{dx}}\int_{ - x}^x {\frac{1}{{1 + t}}} dt = \frac{1}{{1 + x}}\left( 1 \right) - \frac{1}{{1 - x}}\left( { - 1} \right) \cr
& \frac{d}{{dx}}\int_{ - x}^x {\frac{1}{{1 + t}}} dt = \frac{1}{{1 + x}} + \frac{1}{{1 - x}} \cr
& \frac{d}{{dx}}\int_{ - x}^x {\frac{1}{{1 + t}}} dt = \frac{{1 - x + 1 + x}}{{1 - {x^2}}} \cr
& \frac{d}{{dx}}\int_{ - x}^x {\frac{1}{{1 + t}}} dt = \frac{2}{{1 - {x^2}}} \cr} $$
