What is the integral of (1-v)/(v^2-4) ?

The integral of (1-v)/(v^2-4) is -1/4 (ln| v/2+1|-ln| v/2-1|)-1/2 ln|v^2-4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-v)/(v^2-4)

\int \frac{1 - v}{v ^{2} - 4} d v

- \frac{1}{4} (ln \frac{v}{2} + 1 - ln \frac{v}{2} - 1) - \frac{1}{2} ln v^{2} - 4 + C

Solution steps

\int \frac{1 - v}{v ^{2} - 4} d v

Expand

\frac{1 - v}{v ^{2} - 4} : \frac{1}{v ^{2} - 4} - \frac{v}{v ^{2} - 4}

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}{v ^{2} - 4} d v - \int \frac{v}{v ^{2} - 4} d v

\int \frac{1}{v ^{2} - 4} d v = - \frac{1}{4} (ln \frac{v}{2} + 1 - ln \frac{v}{2} - 1)

\int \frac{v}{v ^{2} - 4} d v = \frac{1}{2} ln v^{2} - 4

= - \frac{1}{4} (ln \frac{v}{2} + 1 - ln \frac{v}{2} - 1) - \frac{1}{2} ln v^{2} - 4

Add a constant to the solution

= - \frac{1}{4} (ln \frac{v}{2} + 1 - ln \frac{v}{2} - 1) - \frac{1}{2} ln v^{2} - 4 + C

\int 4 x ln (x) d x

\int x 4 5 x + 1 d x

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

\int x^{4} + x d x

\int x e^{x + 3} d x

What is the integral of (1-v)/(v^2-4) ?
The integral of (1-v)/(v^2-4) is -1/4 (ln| v/2+1|-ln| v/2-1|)-1/2 ln|v^2-4|+C