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

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

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

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

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

Solution steps

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

Complete the square

2 x - x^{2} : - (x - 1)^{2} + 1

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 1} d u

= \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

Substitute back

u = x - 1

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

Simplify

\frac{ln ∣ x - 1 + 1 ∣}{2} - \frac{ln ∣ x - 1 - 1 ∣}{2} : \frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣

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

Add a constant to the solution

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

\int_{- 1}^{0} (x^{3} - x^{2} - 2 x) d x

f (x) = - 3 sin (3 x)

\int \frac{1}{7 x ^{2} - 8} d x

x \to - 3 lim (- x^{2} + 8 x - 9)

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