<s>[INST] Solve the differential equation $2r(1+s^2)dr + (1+r^4)ds = 0$ $2r(1+s^2)dr + (1+r^4)ds = 0$
$\Rightarrow \frac {2r}{1+r^4}dr + \frac {ds}{1+s^2} = 0$ $(1)$
Integrate both sides:
For integration of $\frac {2r}{1+r^4}dr$  $(2)$
I put $1+r^4=t\Rightarrow 4r^3dr=dt\Rightarrow 2rdr = \frac {dt}{2r^2} = \frac {dt}{2\sqrt{t-1}}$
Hence $(2)$ becomes:


*

*Integration of $\frac {dt}{2\sqrt{t-1}}$ = $\sqrt{t-1} = \sqrt{r^4} = r^2$

*Integration of $\frac {ds}{1+s^2} = arctan(s)$


Hence (1) becomes 
$$r^2 + arctan(s) = c$$
But at the back, the answer written is $r^2 + s = c(1-r^2s)$
Can someone please tell me what did I do wrong.
 [/INST] In addition to @Did's hint, there is a relatively easy way to handle that integral. 
We have $$ \frac {2r}{1+ r^4} \mathrm {d}r = - \frac {1}{1+ s^2 } \mathrm {d}s $$ $$\Rightarrow \int \frac {\mathrm {d}(r^2)}{1+(r^2)^2} = \int - \frac {1}{1+ s^2 } \mathrm {d}s$$ $$\Rightarrow \arctan r^2 = - \arctan s +c $$
Hope you can take it from here. 
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