Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 25

$$x=\tan \left(\frac{t^{2}}{2}+t+C\right) $$

Since \begin{aligned} \frac{d x}{d t} &=(t+1)\left(x^{2}+1\right) \\ \frac{d x}{x^{2}+1} &=\frac{(t+1)\left(x^{2}+1\right)}{x^{2}+1} \\ \frac{1}{x^{2}+1} d x &=(t+1) d t \end{aligned} Then \begin{aligned} \int \frac{1}{x^{2}+1} d x &=\int(t+1) d t \\ \tan ^{-1} x &=\frac{t^{2}}{2}+t+C \\ x &=\tan \left(\frac{t^{2}}{2}+t+C\right) \end{aligned}