What is the solution for (dy)/(dx)-tan(x)-1/x =0 ?

The solution for (dy)/(dx)-tan(x)-1/x =0 is y=-ln(cos(x))+ln(x)+c_{1}

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(dy)/(dx)-tan(x)-1/x =0

\frac{d y}{d x} - tan (x) - \frac{1}{x} = 0

y = - ln (cos (x)) + ln (x) + c_{1}

Solution steps

\frac{d y}{d x} - tan (x) - \frac{1}{x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) - tan (x) - \frac{1}{x} = 0

Isolate

y^{'} (x) : y^{'} (x) = tan (x) + \frac{1}{x}

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

then

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

y = \int tan (x) + \frac{1}{x} d x

\int tan (x) + \frac{1}{x} d x = - ln (cos (x)) + ln (x) + c_{1}

y = - ln (cos (x)) + ln (x) + c_{1}

\int \frac{2}{( x ^{2} - 1 ) ^{2}} d x

\int_{0}^{1} \frac{x ^{3} - 5 x - 9}{x ^{2} - x - 6} d x

\frac{1}{x - 2}

f (x) = 3 x^{3 x}

What is the solution for (dy)/(dx)-tan(x)-1/x =0 ?
The solution for (dy)/(dx)-tan(x)-1/x =0 is y=-ln(cos(x))+ln(x)+c_{1}