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

n = 1 \sum \infty \frac{3 n}{n !}

\int csc^{3} (x) \cdot sec (x) d x

\int (x^{2} + a x - 3)^{2} d x

\int_{0}^{1} 2 x^{2} - 1 d x

\frac{d ^{2}}{d x ^{2}} ((x + 1) e^{x})

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