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))

integral

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

\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

\frac{3 x ^{2} + 2 x}{x ^{6}}

1 + 2 x

(1 + 2 cos (x))^{2}

\frac{( x ^{2} - x )}{( x ^{2} + 2 )}

(2 x^{2})^{\frac{1}{3}}

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