What is the solution for f^'(x)=sec(2x)tan(2x)-4/(x^2) ?

The solution for f^'(x)=sec(2x)tan(2x)-4/(x^2) is f(x)= 1/2 cos(2x)+1/2 sin(2x)tan(2x)+4/x+c_{1}

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f^'(x)=sec(2x)tan(2x)-4/(x^2)

f^{^{'}} (x) = sec (2 x) tan (2 x) - \frac{4}{x ^{2}}

f (x) = \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x} + c_{1}

Solution steps

f^{'} (x) = sec (2 x) tan (2 x) - \frac{4}{x ^{2}}

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

then

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

f (x) = \int sec (2 x) tan (2 x) - \frac{4}{x ^{2}} d x

\int sec (2 x) tan (2 x) - \frac{4}{x ^{2}} d x = \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x} + c_{1}

f (x) = \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x} + c_{1}

What is the solution for f^'(x)=sec(2x)tan(2x)-4/(x^2) ?
The solution for f^'(x)=sec(2x)tan(2x)-4/(x^2) is f(x)= 1/2 cos(2x)+1/2 sin(2x)tan(2x)+4/x+c_{1}