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Popular >

integral of (1/y-1)

Solution

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

Solution

ln ∣ y ∣ - y + C

Solution steps

\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

Popular Examples

y^{''}+4y^'+6y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 4 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'}} = x + y, y (0) = 9

derivative of x+1-tan(x/4)

\frac{d}{d x} (x + 1 - tan (\frac{x}{4}))

limit as x approaches 6-of 1/(x-6)

x \to 6 - lim (\frac{1}{x - 6})

integral of 9/((x-4)(x+2))

\int \frac{9}{( x - 4 ) ( x + 2 )} d x

Frequently Asked Questions (FAQ)

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