What is the integral of 1/(p-z)+1/z ?

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

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integral of 1/(p-z)+1/z

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

- ln ∣ p - z ∣ + ln ∣ z ∣ + C

Solution steps

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

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

\int \frac{1}{p - z} d z = - ln ∣ p - z ∣

\int \frac{1}{z} d z = ln ∣ z ∣

= - ln ∣ p - z ∣ + ln ∣ z ∣

Add a constant to the solution

= - ln ∣ p - z ∣ + ln ∣ z ∣ + C

\int 4 (x - 1) d x

\frac{x ^{6}}{2 - x ^{5}}

f (x) = \frac{5}{4} x^{\frac{4}{3}}

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

\frac{x ^{3}}{x - 2}

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