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

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

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integral of ((1-x)y)/x

\int \frac{( 1 - x ) y}{x}

y (ln ∣ x ∣ - x) + C

Solution steps

\int \frac{( 1 - x ) y}{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= y \cdot \int \frac{1 - x}{x} d x

Expand

\frac{1 - x}{x} : \frac{1}{x} - 1

= y \cdot \int \frac{1}{x} - 1 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

= y (\int \frac{1}{x} d x - \int 1 d x)

\int \frac{1}{x} d x = ln ∣ x ∣

\int 1 d x = x

= y (ln ∣ x ∣ - x)

Add a constant to the solution

= y (ln ∣ x ∣ - x) + C

x \to 0 lim (∣ 5 x ∣)

\frac{d}{d x} (\frac{1}{3 2 x - 1})

\frac{d}{d x} (x ln (ln (x)))

\int \frac{x ^{2} + 1}{( x + 2 ) ( x ^{2} + 2 x + 5 )} d x

\frac{d y}{d x}, 2 x y = 5