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Popular >

integral of z/(z-1)

Solution

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

Solution

z - 1 + ln ∣ z - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 1}{u} d u

= \int \frac{u + 1}{u} d u

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

= \int 1 + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d u + \int \frac{1}{u} d u

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u + ln ∣ u ∣

Substitute back

u = z - 1

= z - 1 + ln ∣ z - 1 ∣

Add a constant to the solution

= z - 1 + ln ∣ z - 1 ∣ + C

Popular Examples

integral of 2/(x^2-x)

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

\frac{d v}{d t}

(x+y)^2dx+(2xy+x^2-3)dy=0

(x + y)^{2} d x + (2 x y + x^{2} - 3) d y = 0

(\partial)/(\partial x)((e^x)^2)

\frac{\partial}{\partial x} ((e^{x})^{2})

limit as x approaches c of-6+3(1/3)

x \to c lim (- 6 + 3 (\frac{1}{3}))

Frequently Asked Questions (FAQ)

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