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

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

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

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

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

Solution steps

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

Complete the square

x (2 - x) : - (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 \frac{1}{6 x - 1} d x

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sin (t + π)

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

y = 9 e^{x} cos (x)

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