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

The integral of (3-z)/(-1+z^2) is -3(1/2 ln|z+1|-1/2 ln|z-1|)-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{1}{2} ln ∣ z + 1 ∣ - \frac{1}{2} ln ∣ z - 1 ∣) - \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{1}{2} ln ∣ z + 1 ∣ - \frac{1}{2} ln ∣ z - 1 ∣)

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

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

Add a constant to the solution

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

cos (sin (5) x^{3})

\frac{d}{d x} (ln (\frac{x}{a}))

\int \frac{8 + t + t ^{2}}{t} d t

\int e^{3 x} d x

8 x^{2} + 3 x

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