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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\int \frac{2 + x}{x} d x

\frac{d}{d x} (e^{a a r c c o s (x)})

2 cos (9 x), 2 - 2 cos (9 x), [0, \frac{π}{9}]

\int 6 arctan (x) d x

x \to 8 + lim (\frac{1}{∣ 8 - x ∣})

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