What is the integral of (sec(2x)tan(2x)-4/(x^2)) ?

The integral of (sec(2x)tan(2x)-4/(x^2)) is 1/2 cos(2x)+1/2 sin(2x)tan(2x)+4/x+C

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integral of (sec(2x)tan(2x)-4/(x^2))

\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

Solution steps

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

Apply the Sum Rule:

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

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

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

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

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

Simplify

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

Add a constant to the solution

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

\int \frac{x - 2}{( x - 3 ) ( x - 4 )} d x

\int \frac{1}{t ^{3}} d t

x \to 1 + lim (\frac{1 + x}{x - 1})

x \to - \infty lim (5 x - 2^{x})

\frac{\partial}{\partial x} (\frac{9 y}{x ^{5}})

What is the integral of (sec(2x)tan(2x)-4/(x^2)) ?
The integral of (sec(2x)tan(2x)-4/(x^2)) is 1/2 cos(2x)+1/2 sin(2x)tan(2x)+4/x+C