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

The integral of 1/(1-2x-3x^2) is -1/4 ln|3x-1|+1/4 ln|x+1|+C

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integral of 1/(1-2x-3x^2)

\int \frac{1}{1 - 2 x - 3 x ^{2}} d x

- \frac{1}{4} ln ∣ 3 x - 1 ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{1}{1 - 2 x - 3 x ^{2}} d x

Take the partial fraction of

\frac{1}{1 - 2 x - 3 x ^{2}} : - \frac{3}{4 ( 3 x - 1 )} + \frac{1}{4 ( x + 1 )}

= \int - \frac{3}{4 ( 3 x - 1 )} + \frac{1}{4 ( 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{3}{4 ( 3 x - 1 )} d x + \int \frac{1}{4 ( x + 1 )} d x

\int \frac{3}{4 ( 3 x - 1 )} d x = \frac{1}{4} ln ∣ 3 x - 1 ∣

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

= - \frac{1}{4} ln ∣ 3 x - 1 ∣ + \frac{1}{4} ln ∣ x + 1 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ 3 x - 1 ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + C

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What is the integral of 1/(1-2x-3x^2) ?
The integral of 1/(1-2x-3x^2) is -1/4 ln|3x-1|+1/4 ln|x+1|+C