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

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

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

integral

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

\frac{x ^{2}}{2} - 2 x + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int 2 d x = 2 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{x ^{2}}{2} - 2 x + ln ∣ x ∣

Add a constant to the solution

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

x^{2} + 4

x^{2} + 3

(20 - 2 x)^{\frac{1}{2}}

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

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

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