Find general solution of ODE $(1+t^2)x''+(x')^2+1=0$ I want to solve following ODE:
$$
(1+t^2)x''+(x')^2+1=0
$$
substitute $y = x'$
$$(1+t^2)y' + y^2 + 1 = 0\\
\frac{dy}{dt}=\frac{-(1+y^2)}{1+t^2}\\
\frac{dy}{1+y^2} = \frac{-dt}{1+t^2}
$$
integrating both sides leaves me with
$$
\arctan y = - \arctan t + c
$$
taking $\tan$ from both sides
$$
y = \tan ( - \arctan t + c)\\
\frac{dx}{dt} = \tan ( - \arctan t + c)\\
dx = dt \tan ( - \arctan t + c)
$$
I surely can't integrate that - I must have done mistake somewhere. Could somebody point me where and how to fix it?

By reassigning the constant
$$\arctan y = \arctan C_1 - \arctan t = \arctan\left(\frac{C_1-t}{1+C_1t}\right)$$
one can take tangent on both sides and integrate getting the solution
$$x = \left(1+\frac{1}{C_1^2}\right)\log\left(1+C_1t\right) - \frac{1}{C_1^2}t + C_2$$
or in the case $C_1 \to \infty$ and $C_1 = 0$ we have
$$x = \log t+C_2$$
$$x = -\frac{1}{2}t^2+C_2$$
respectively.
