Answer
$$x = \tan \left( {\ln \left| {t + 1} \right|} \right)$$
Work Step by Step
$$\eqalign{
& \left( {t + 1} \right)\frac{{dx}}{{dt}} = {x^2} + 1 \cr
& {\text{Separating the variables}} \cr
& \frac{{dx}}{{{x^2} + 1}} = \frac{{dt}}{{t + 1}} \cr
& \cr
& {\text{Integrate}} \cr
& \int {\frac{{dx}}{{{x^2} + 1}}} = \int {\frac{{dt}}{{t + 1}}} \cr
& \arctan x = \ln \left| {t + 1} \right| + C \cr
& \cr
& {\text{Use the condition }}x\left( 0 \right) = 0 \cr
& \arctan 0 = \ln \left| {0 + 1} \right| + C \cr
& C = 0 \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \arctan x = \ln \left| {t + 1} \right| \cr
& {\text{Solve for }}x \cr
& \tan \left( {\arctan x} \right) = \tan \left( {\ln \left| {t + 1} \right|} \right) \cr
& x = \tan \left( {\ln \left| {t + 1} \right|} \right) \cr} $$
