Answer
$$ = \frac{1}{2}ta{n^{ - 1}}\left( {{t^2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{t}{{{t^4} + 1}}} dt \cr
& or \cr
& \int {\frac{t}{{{{\left( {{t^2}} \right)}^2} + 1}}} dt \cr
& {\text{substitutue }}u = {t^2},{\text{ }}du = 2tdt,{\text{ }}tdt = \frac{1}{2}du \cr
& \int {\frac{t}{{{{\left( {{t^2}} \right)}^2} + 1}}} dt = \int {\frac{{\left( {1/2} \right)du}}{{{u^2} + 1}}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{{u^2} + 1}}} \cr
& {\text{ use the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}ta{n^{ - 1}}\left( {\frac{u}{a}} \right) + C,{\text{ }}\left( {{\text{see page 468}}} \right) \cr
& = \frac{1}{2}\left( {\frac{1}{1}ta{n^{ - 1}}\left( {\frac{u}{1}} \right)} \right) + C \cr
& = \frac{1}{2}ta{n^{ - 1}}\left( u \right) + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}ta{n^{ - 1}}\left( {{t^2}} \right) + C \cr} $$
