<s>[INST] reduce a differential equation $y^{'}=\dfrac{1+y}{1-x}$ I want to reduce a differential equation.
$$y^{'}=\dfrac{1+y}{1-x}$$
I reduce this, but my answer don't much "Wolfram alpha".
Please tell me what is wrong.

divide both sides of the equality by $(1+y)$
$$\dfrac{y'}{1+y}=\dfrac{1}{1-x}$$
integrate both sides by $x$
$$\int \dfrac{1}{1+y}dy=\int \dfrac{1}{1-x}dx$$
$$\log \left| 1+y\right| =-\log \left| 1-x\right| +C$$
$$1+y=\dfrac{C}{1-x}$$
Last
$$y=\dfrac{C}{1-x}-\dfrac{1}{1-x}$$
 [/INST] From your calculations
$$1+y = \dfrac C {1-x} \implies y = \dfrac C{1-x} - 1$$
which is already the solution.
If you want to obtain the form given by WolframAlpha, notice that:
$$\frac C{1-x} - 1 = \frac C{1-x}-\frac 1{1-x} + \frac x{1-x} = \frac{C-1}{1-x}+\frac x{1-x}$$
and $C-1$ is absorbed into $C$ since $C$ can be any constant, giving
$$y = \frac {c_1}{1-x}+\frac x{1-x}$$
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