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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(1-x-2x^2)

\int \frac{1}{1 - x - 2 x ^{2}} d x

- \frac{1}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{1}{1 - x - 2 x ^{2}} d x

Take the partial fraction of

\frac{1}{1 - x - 2 x ^{2}} : - \frac{2}{3 ( 2 x - 1 )} + \frac{1}{3 ( x + 1 )}

= \int - \frac{2}{3 ( 2 x - 1 )} + \frac{1}{3 ( x + 1 )} d x

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{2}{3 ( 2 x - 1 )} d x + \int \frac{1}{3 ( x + 1 )} d x

\int \frac{2}{3 ( 2 x - 1 )} d x = \frac{1}{3} ln ∣ 2 x - 1 ∣

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

= - \frac{1}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x + 1 ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + C

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