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

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

Popular >

integral of (1-y)/y

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

ln ∣ y ∣ - y + C

Solution steps

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

Expand

\frac{1 - y}{y} : \frac{1}{y} - 1

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

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}{y} d y - \int 1 d y

\int \frac{1}{y} d y = ln ∣ y ∣

\int 1 d y = y

= ln ∣ y ∣ - y

Add a constant to the solution

= ln ∣ y ∣ - y + C

\int_{0}^{2.1} π (\frac{21}{10} - x)^{2} d x

\frac{\partial ^{2}}{\partial y ^{2}} (ln (2 + x^{2} y^{2}))

\int \frac{1}{9} x^{- \frac{2}{9}} d x

\frac{\partial}{\partial x} (y^{2} \cdot x)

\int_{1}^{9} \frac{1}{2 x ^{2}} d x