<s>[INST] Solve $\frac{dx}{dt} = x^3 + x$ for $x$ This is a seemingly simple first order separable differential equation that I'm getting stuck on.  This is what I have so far:
$$\frac{dx}{dt} = x^3+x$$
goes to
$$\frac{dx}{x(1+x^2)} = dt$$
Now using partial fractions to integrate the left-hand side:
$$\frac{1}{x(1+x^2)} = \frac{A}{x} + \frac{Bx+C}{1+x^2}$$
Solving for A, B, C:
$1 = A(1+x^2) + (Bx+C)x$, and using coefficient matching, I get $A=1, B=-1, C=0$.
So the integral yields: 
$$\int\frac{1}{x} - \frac{x}{1+x^2}dx = \int dt$$
This yields: $$\ln x -\frac{1}{2}\ln(1+x^2) =t + C$$
So I tried using log rules and such to solve for $x$.  I think this is where my source of error is.
My attempt:
$$\ln x - \ln(1+x^2)^\frac{1}{2} = t+C$$
$$\ln\frac{x}{(1+x^2)^{\frac{1}{2}}} = t + C$$
$$\frac{x}{(1+x^2)^{\frac{1}{2}}}= Ce^t$$
But this seems wrong.  I apologize in advance if it's a silly mistake that I didn't see.
 [/INST] Assume that $x\ge 0$.
$$\frac{x}{(1+x^2)^{\frac{1}{2}}}= \frac{\sqrt{x^2}}{(1+x^2)^{\frac{1}{2}}}=\sqrt{\frac{x^2}{1+x^2}}$$
and you can probably take it from there.
If $x<0$, then $x=-\sqrt{x^2}$ and proceed similarly.
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