What is the integral of ((1-2x)^2)/x ?

The integral of ((1-2x)^2)/x is ln|x|-4x+2x^2+C

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integral of ((1-2x)^2)/x

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

ln ∣ x ∣ - 4 x + 2 x^{2} + C

Solution steps

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

Expand

\frac{( 1 - 2 x ) ^{2}}{x} : \frac{1}{x} - 4 + 4 x

= \int \frac{1}{x} - 4 + 4 x 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{1}{x} d x - \int 4 d x + \int 4 x d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int 4 d x = 4 x

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

= ln ∣ x ∣ - 4 x + 2 x^{2}

Add a constant to the solution

= ln ∣ x ∣ - 4 x + 2 x^{2} + C

\frac{d}{d x} (- x^{2} + 25)

y d x + x^{\frac{α}{6}} d y = 8 y x d x + 2 d y

t a y l or ln (x^{2} + 1)

d er i v a t i v e - 6 t (5 t^{3} - 1)^{7}

x \to - 3 lim (x^{2} - 5)

What is the integral of ((1-2x)^2)/x ?
The integral of ((1-2x)^2)/x is ln|x|-4x+2x^2+C