[tex]\displaystyle
\dfrac{dx}{dt}=2(x^2+1)\\\\
\int_{t_0}^t dt=\int_{x_0}^x\dfrac{dx}{2(x^2+1)}\\\\
t-t_0=\dfrac{1}{2}\int_{x_0}^x\dfrac{dx}{(x^2+1)}\\\\
t-t_0=\dfrac{1}{2}\left[\arctan(x)\right]_{x_0}^x\\\\
t-t_0=\dfrac{1}{2}\left[\arctan(x)-\arctan(x_0)\right][/tex]
We'll use [tex]x(\pi/4)=1[/tex], considering that [tex]x_0=1, t_0=\dfrac{\pi}{4}[/tex]:
[tex]t-t_0=\dfrac{1}{2}\left[\arctan(x)-\arctan(x_0)\right]\\\\
t-\dfrac{\pi}{4}=\dfrac{1}{2}\left[\arctan(x)-\arctan(1)\right]\\\\
t-\dfrac{\pi}{4}=\dfrac{1}{2}\left[\arctan(x)-\dfrac{\pi}{4}\right]\\\\
t-\dfrac{\pi}{4}=\dfrac{1}{2}\arctan(x)-\dfrac{\pi}{8}\\\\
t-\dfrac{\pi}{8}=\dfrac{1}{2}\arctan(x)\\\\
\boxed{\arctan(x)=2t-\dfrac{\pi}{4}}[/tex]
Applying tan in the both sides:
[tex]\arctan(x)=2t-\dfrac{\pi}{4}\\\\
\tan(\arctan(x))=\tan\left(2t-\dfrac{\pi}{4}\right)\\\\
\boxed{x(t)=\tan\left(2t-\dfrac{\pi}{4}\right)}[/tex]
