What is the solution for (dy)/(dx)=(x-1)/(3x^2-6x-1) ?

The solution for (dy)/(dx)=(x-1)/(3x^2-6x-1) is y= 1/6 ln(3x^2-6x-1)+c_{1}

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(dy)/(dx)=(x-1)/(3x^2-6x-1)

\frac{d y}{d x} = \frac{x - 1}{3 x ^{2} - 6 x - 1}

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

Solution steps

\frac{d y}{d x} = \frac{x - 1}{3 x ^{2} - 6 x - 1}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{x - 1}{3 x ^{2} - 6 x - 1}

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

then

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

y = \int \frac{x - 1}{3 x ^{2} - 6 x - 1} d x

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

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

\int (ln (x)) 2 d x

\frac{ln ( x )}{ln ( 2 )}

\int \frac{4 x ^{2} + 9}{x ^{4}} d x

\int \frac{x}{9 + x ^{4}} d x

\frac{1 + 0.5 x}{1 - 0.5 x}

What is the solution for (dy)/(dx)=(x-1)/(3x^2-6x-1) ?
The solution for (dy)/(dx)=(x-1)/(3x^2-6x-1) is y= 1/6 ln(3x^2-6x-1)+c_{1}