Answer
$$4{\sec ^{ - 1}}x + \frac{{{x^2}}}{2} + {\tan ^{ - 1}}x + C$$
Work Step by Step
$$\eqalign{
& \int {\left[ {\frac{4}{{x\sqrt {{x^2} - 1} }} + \frac{{1 + x + {x^3}}}{{1 + {x^2}}}} \right]} dx \cr
& {\text{Split the integrand}} \cr
& = \int {\frac{4}{{x\sqrt {{x^2} - 1} }}} dx + \int {\frac{{1 + x + {x^3}}}{{1 + {x^2}}}} dx \cr
& {\text{By long division }}\frac{{1 + x + {x^3}}}{{1 + {x^2}}} = x + \frac{1}{{{x^2} + 1}} \cr
& = \int {\frac{4}{{x\sqrt {{x^2} - 1} }}} dx + \int {\left( {x + \frac{1}{{{x^2} + 1}}} \right)} dx \cr
& = 4\int {\frac{1}{{x\sqrt {{x^2} - 1} }}} dx + \int x dx + \int {\frac{1}{{{x^2} + 1}}} dx \cr
& {\text{Use the rule of integration for inverse trigonometric functions}} \cr
& = 4{\sec ^{ - 1}}x + \frac{{{x^2}}}{2} + {\tan ^{ - 1}}x + C \cr
& \cr
& \cr
& {\text{Check by differentiation}} \cr
& \frac{d}{{dx}}\left[ {4{{\sec }^{ - 1}}x + \frac{{{x^2}}}{2} + {{\tan }^{ - 1}}x + C} \right] \cr
& = 4\left( {\frac{1}{{x\sqrt {{x^2} - 1} }}} \right) + \frac{{2x}}{2} + \frac{1}{{1 + {x^2}}} + 0 \cr
& = \frac{4}{{x\sqrt {{x^2} - 1} }} + x + \frac{1}{{1 + {x^2}}} \cr
& = \frac{4}{{x\sqrt {{x^2} - 1} }} + \frac{{x + {x^3} + 1}}{{1 + {x^2}}} \cr} $$
