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

The integral of ((2-4x)^2)/(x^2) is -4/x-16ln|x|+16x+C

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

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

- \frac{4}{x} - 16 ln ∣ x ∣ + 16 x + C

Solution steps

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

Expand

\frac{( 2 - 4 x ) ^{2}}{x ^{2}} : \frac{4}{x ^{2}} - \frac{16}{x} + 16

= \int \frac{4}{x ^{2}} - \frac{16}{x} + 16 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 \frac{4}{x ^{2}} d x - \int \frac{16}{x} d x + \int 16 d x

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

\int \frac{16}{x} d x = 16 ln ∣ x ∣

\int 16 d x = 16 x

= - \frac{4}{x} - 16 ln ∣ x ∣ + 16 x

Add a constant to the solution

= - \frac{4}{x} - 16 ln ∣ x ∣ + 16 x + C

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What is the integral of ((2-4x)^2)/(x^2) ?
The integral of ((2-4x)^2)/(x^2) is -4/x-16ln|x|+16x+C