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y =tan[C- arctan(x)]
The original equation is separable, so you should get:
Integral of [(1/(1 + y^2))dy] = Integral of [(1/(1 + x^2))dx]
which integrates to give:
arctan(y) = - arctan(x) + C
[where arctan(x) = tan^-1(x)]
Thus, on taking the tangent of both sides, we have
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