What is the integral of (3+z)/(1-z^2) ?

The integral of (3+z)/(1-z^2) is 3((ln|z+1|)/2-(ln|z-1|)/2)-1/2 ln|1-z^2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3+z)/(1-z^2)

\int \frac{3 + z}{1 - z ^{2}} d z

3 (\frac{ln ∣ z + 1 ∣}{2} - \frac{ln ∣ z - 1 ∣}{2}) - \frac{1}{2} ln 1 - z^{2} + C

Solution steps

\int \frac{3 + z}{1 - z ^{2}} d z

Expand

\frac{3 + z}{1 - z ^{2}} : \frac{3}{1 - z ^{2}} + \frac{z}{1 - z ^{2}}

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{3}{1 - z ^{2}} d z + \int \frac{z}{1 - z ^{2}} d z

\int \frac{3}{1 - z ^{2}} d z = 3 (\frac{ln ∣ z + 1 ∣}{2} - \frac{ln ∣ z - 1 ∣}{2})

\int \frac{z}{1 - z ^{2}} d z = - \frac{1}{2} ln 1 - z^{2}

= 3 (\frac{ln ∣ z + 1 ∣}{2} - \frac{ln ∣ z - 1 ∣}{2}) - \frac{1}{2} ln 1 - z^{2}

Add a constant to the solution

= 3 (\frac{ln ∣ z + 1 ∣}{2} - \frac{ln ∣ z - 1 ∣}{2}) - \frac{1}{2} ln 1 - z^{2} + C

n = 0 \sum \infty \frac{1}{n ^{2}} x^{n}

y = \frac{2}{3} x^{\frac{3}{2}} + 3, [0, 2]

y = x^{n}

\frac{6}{y} d x = (e^{3 x} + 1) d y

(x + 1) (\frac{d y}{d x}) = x + 6

What is the integral of (3+z)/(1-z^2) ?
The integral of (3+z)/(1-z^2) is 3((ln|z+1|)/2-(ln|z-1|)/2)-1/2 ln|1-z^2|+C