What is the integral of (2x^2cos(4x)-3)/(x^2) ?

The integral of (2x^2cos(4x)-3)/(x^2) is 1/2 sin(4x)+3/x+C

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integral of (2x^2cos(4x)-3)/(x^2)

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

\frac{1}{2} sin (4 x) + \frac{3}{x} + C

Solution steps

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

Expand

\frac{2 x ^{2} cos ( 4 x ) - 3}{x ^{2}} : 2 cos (4 x) - \frac{3}{x ^{2}}

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

\int 2 cos (4 x) d x = \frac{1}{2} sin (4 x)

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

= \frac{1}{2} sin (4 x) - (- \frac{3}{x})

Simplify

= \frac{1}{2} sin (4 x) + \frac{3}{x}

Add a constant to the solution

= \frac{1}{2} sin (4 x) + \frac{3}{x} + C

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What is the integral of (2x^2cos(4x)-3)/(x^2) ?
The integral of (2x^2cos(4x)-3)/(x^2) is 1/2 sin(4x)+3/x+C