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

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

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

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

ln ∣ x ∣ - 2 x + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

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

= \int \frac{1}{x} - 2 + 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 2 d x + \int x d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int 2 d x = 2 x

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

= ln ∣ x ∣ - 2 x + \frac{x ^{2}}{2}

Add a constant to the solution

= ln ∣ x ∣ - 2 x + \frac{x ^{2}}{2} + C

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

\int 3 x^{2} - 2 x + 1 - \frac{1}{x ^{2}} d x

\int x - 2 ∣ x ∣ d x

\int - 5 x (4 - e^{x^{2}}) d x

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

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