Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{Cannot be evaluated with the basic integration}} \cr
& {\text{rules studied so far}}{\text{.}} \cr
& \left( {\text{b}} \right)\frac{1}{2}\arctan \left( {{x^2}} \right) + C \cr
& \left( {\text{c}} \right)\ln \left( {1 + {x^4}} \right) + C \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\int {\frac{1}{{1 + {x^4}}}} dx \cr
& {\text{This integral cannot be evaluated with the basic integration}} \cr
& {\text{rules studied so far}}{\text{.}} \cr
& \cr
& \left( {\text{b}} \right)\int {\frac{x}{{1 + {x^4}}}} dx \cr
& = \int {\frac{x}{{1 + {{\left( {{x^2}} \right)}^2}}}} dx \cr
& {\text{Integrate by substitution, let }}u = {x^2},{\text{ }}du = 2xdx \cr
& \int {\frac{x}{{1 + {{\left( {{x^2}} \right)}^2}}}} dx = \frac{1}{2}\int {\frac{{du}}{{1 + {u^2}}}} \cr
& = \frac{1}{2}\arctan \left( {\frac{u}{a}} \right) + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}\arctan \left( {{x^2}} \right) + C \cr
& \cr
& \left( {\text{c}} \right)\int {\frac{{{x^3}}}{{1 + {x^4}}}} dx \cr
& {\text{Integrate by substitution, let }}u = 1 + {x^4},{\text{ }}du = 4{x^3}dx \cr
& \int {\frac{{{x^3}}}{{1 + {x^4}}}} dx = \frac{1}{4}\int {\frac{{du}}{u}} \cr
& = \ln \left| u \right| + C \cr
& {\text{write in terms of }}x \cr
& = \ln \left( {1 + {x^4}} \right) + C \cr} $$
