What is the solution for 6x^2+(x^3+2) derivative of y=0 ?

The solution for 6x^2+(x^3+2) derivative of y=0 is y=-2ln(x^3+2)+c_{1}

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6x^2+(x^3+2) derivative of y=0

6 x^{2} + (x^{3} + 2) \frac{d}{d x} y = 0

y = - 2 ln (x^{3} + 2) + c_{1}

Solution steps

6 x^{2} + (x^{3} + 2) \frac{d y}{d x} = 0

Isolate

y^{'} (x) : y^{'} (x) = - \frac{6 x ^{2}}{x ^{3} + 2}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - \frac{6 x ^{2}}{x ^{3} + 2} d x

\int - \frac{6 x ^{2}}{x ^{3} + 2} d x = - 2 ln (x^{3} + 2) + c_{1}

y = - 2 ln (x^{3} + 2) + c_{1}

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What is the solution for 6x^2+(x^3+2) derivative of y=0 ?
The solution for 6x^2+(x^3+2) derivative of y=0 is y=-2ln(x^3+2)+c_{1}