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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

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