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

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

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x y^{^{'}} + 2 y = x^{3} - 4

\int_{4}^{\infty} 7 e^{- \frac{x}{2}} d x

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

\frac{d}{d x} (4 cos (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