\(\begin{array}{l}
1)\quad x(1 + x^2)yy' - (1 + y^2) =0\\
\Leftrightarrow x(1 + x^2)y\cdot \dfrac{dy}{dx} - (1 + y^2) = 0\\
\Leftrightarrow x(1 + x^2)ydy - (1+y^2)dx = 0\\
\Leftrightarrow \dfrac{y}{1 + y^2}dy = \dfrac{1}{x(1 + x^2)}dx\\
\Leftrightarrow \displaystyle\int\dfrac{y}{1+y^2} = \displaystyle\int\dfrac{1}{x(1+x^2)}dx\\
\Leftrightarrow \dfrac12\ln(1 + y^2) = \ln x - \dfrac12\ln(1 + x^2) + C\\
\Leftrightarrow \ln(1 + y^2) = 2\ln x - \ln(1 + x^2) + C\\
\Leftrightarrow \ln(1 + y^2) = \ln\dfrac{x^2}{1 + x^2} + C\\
\Leftrightarrow 1 + y^2 = C\cdot \dfrac{x^2}{1 + x^2}\\
\Leftrightarrow y^2 = C\cdot \dfrac{x^2}{1 + x^2} -1\\
\Leftrightarrow y = \pm \sqrt{C\cdot \dfrac{x^2}{1 + x^2} - 1}\\
2)\quad \quad (1+y^2)dx - 4x(1+x)dy = 0\\
\Leftrightarrow \dfrac{4}{y^2+1}dy = \dfrac{1}{x(x+1)}dx\\
\Leftrightarrow \displaystyle\int\dfrac{4}{y^2+1}dy = \displaystyle\int\dfrac{1}{x(x+1)}dx\\
\Leftrightarrow 4\arctan y = \ln x - \ln(x+1) + C\\
\Leftrightarrow y = \tan\left(\dfrac14\ln\dfrac{x}{x+1} + C\right)
\end{array}\)
