What is the integral of 1/(y^2-4y-12) ?

The integral of 1/(y^2-4y-12) is -1/8 ln|y+2|+1/8 ln|y-6|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(y^2-4y-12)

\int \frac{1}{y ^{2} - 4 y - 12} d y

- \frac{1}{8} ln ∣ y + 2 ∣ + \frac{1}{8} ln ∣ y - 6 ∣ + C

Solution steps

\int \frac{1}{y ^{2} - 4 y - 12} d y

Take the partial fraction of

\frac{1}{y ^{2} - 4 y - 12} : - \frac{1}{8 ( y + 2 )} + \frac{1}{8 ( y - 6 )}

= \int - \frac{1}{8 ( y + 2 )} + \frac{1}{8 ( y - 6 )} d y

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}{8 ( y + 2 )} d y + \int \frac{1}{8 ( y - 6 )} d y

\int \frac{1}{8 ( y + 2 )} d y = \frac{1}{8} ln ∣ y + 2 ∣

\int \frac{1}{8 ( y - 6 )} d y = \frac{1}{8} ln ∣ y - 6 ∣

= - \frac{1}{8} ln ∣ y + 2 ∣ + \frac{1}{8} ln ∣ y - 6 ∣

Add a constant to the solution

= - \frac{1}{8} ln ∣ y + 2 ∣ + \frac{1}{8} ln ∣ y - 6 ∣ + C

\int \frac{1}{( 2 x ^{2} ) ^{2}} d x

\int e^{- a x} cos (x) d x

\int x e^{- 3} d x

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

\int sin^{4} (\frac{t}{2}) d t

What is the integral of 1/(y^2-4y-12) ?
The integral of 1/(y^2-4y-12) is -1/8 ln|y+2|+1/8 ln|y-6|+C