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

The integral of 1/(2x^2-x-3) is -1/5 ln|x+1|+1/5 ln|2x-3|+C

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

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

- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ 2 x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{2 x ^{2} - x - 3} : - \frac{1}{5 ( x + 1 )} + \frac{2}{5 ( 2 x - 3 )}

= \int - \frac{1}{5 ( x + 1 )} + \frac{2}{5 ( 2 x - 3 )} 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{1}{5 ( x + 1 )} d x + \int \frac{2}{5 ( 2 x - 3 )} d x

\int \frac{1}{5 ( x + 1 )} d x = \frac{1}{5} ln ∣ x + 1 ∣

\int \frac{2}{5 ( 2 x - 3 )} d x = \frac{1}{5} ln ∣ 2 x - 3 ∣

= - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ 2 x - 3 ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ 2 x - 3 ∣ + C

\int sin^{7} (\frac{3}{5} x) d x

\int \frac{x}{x + 9} d x

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

y = 3 x^{2} - 6, (- 1, - 3)

\int_{2}^{\infty} e^{- 9 p} d p

What is the integral of 1/(2x^2-x-3) ?
The integral of 1/(2x^2-x-3) is -1/5 ln|x+1|+1/5 ln|2x-3|+C