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

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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{- 2 x ^{2} - x + 1} : - \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

y^{^{''}} - y^{^{'}} = 1

\int \frac{4 ^{x} + 1 0 ^{x}}{2 ^{x}} d x

\int \frac{1}{x ( 1 + x )} d x

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

\frac{d}{d x} (cos^{3} (x^{2} - 1))

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