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

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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

\int_{- 125}^{64} \frac{1}{x ^{\frac{1}{3}}} d x

\frac{\partial}{\partial y} (x y z^{2})

x \to \infty lim (- 0. 5^{x})

\int \frac{4 x ^{4} + 3}{4 x ^{5} + 15 x + 2} d x

ln (sin (4 x))

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