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

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

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

\int \frac{1 - 2 y}{y ^{2}} d y

- \frac{1 - 2 y}{y} - 2 ln ∣ y ∣ + C

Solution steps

\int \frac{1 - 2 y}{y ^{2}} d y

Apply Integration By Parts

: - \frac{1 - 2 y}{y} - \int \frac{2}{y} d y

= - \frac{1 - 2 y}{y} - \int \frac{2}{y} d y

\int \frac{2}{y} d y = 2 ln ∣ y ∣

= - \frac{1 - 2 y}{y} - 2 ln ∣ y ∣

Add a constant to the solution

= - \frac{1 - 2 y}{y} - 2 ln ∣ y ∣ + C

\frac{d}{d x} (e^{x + y} - x)

\frac{d}{d x} (cos (sin (6 x)))

\frac{\partial}{\partial y} (\frac{1 + t ^{2}}{3 y - y ^{2}})

\frac{4 x ^{2} + 6 x + 6}{x}

x \to \frac{π}{2} lim (\frac{1}{cos ( x )})

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