What is the integral of ((y-3))/(4y) ?

The integral of ((y-3))/(4y) is 1/4 (y-3ln|y|)+C

Popular >

integral of ((y-3))/(4y)

\int \frac{( y - 3 )}{4 y} d y

\frac{1}{4} (y - 3 ln ∣ y ∣) + C

Solution steps

\int \frac{y - 3}{4 y} d y

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{y - 3}{y} d y

Expand

\frac{y - 3}{y} : 1 - \frac{3}{y}

= \frac{1}{4} \cdot \int 1 - \frac{3}{y} 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

= \frac{1}{4} (\int 1 d y - \int \frac{3}{y} d y)

\int 1 d y = y

\int \frac{3}{y} d y = 3 ln ∣ y ∣

= \frac{1}{4} (y - 3 ln ∣ y ∣)

Add a constant to the solution

= \frac{1}{4} (y - 3 ln ∣ y ∣) + C

\int 5 x^{- 3} + 8 x^{- 2} + 2 x^{- 1} d x

\int e^{- x} sin (x) + e^{- x} cos (x) d x

\int cos (x + 3) d x

\int - \frac{1}{8} cos (8 x) d x

What is the integral of ((y-3))/(4y) ?
The integral of ((y-3))/(4y) is 1/4 (y-3ln|y|)+C