Answer
$$12{\sec ^{ - 1}}x + x - \frac{{{x^3}}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {\frac{{12}}{{x\sqrt {{x^2} - 1} }} + \frac{{1 - {x^4}}}{{1 + {x^2}}}} \right)} dx \cr
& {\text{sum rule}} \cr
& = \int {\frac{{12}}{{x\sqrt {{x^2} - 1} }}} dx + \int {\frac{{1 - {x^4}}}{{1 + {x^2}}}} dx \cr
& {\text{factor difference of squares}} \cr
& = \int {\frac{{12}}{{x\sqrt {{x^2} - 1} }}} dx + \int {\frac{{\left( {1 + {x^2}} \right)\left( {1 - {x^2}} \right)}}{{1 + {x^2}}}} dx \cr
& = 12\int {\frac{1}{{x\sqrt {{x^2} - 1} }}} dx + \int {\left( {1 - {x^2}} \right)} dx \cr
& {\text{find the antiderivative}} \cr
& = 12\left( {{{\sec }^{ - 1}}x} \right) + x - \frac{{{x^{2 + 1}}}}{{2 + 1}} + C \cr
& = 12{\sec ^{ - 1}}x + x - \frac{{{x^3}}}{3} + C \cr} $$