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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\frac{\partial}{\partial t} (e^{- 7 t} cos (π x))

x \to \frac{π}{4} lim (\frac{sin ( x - \frac{π}{4} )}{4 x - π})

x^{^{'}} + x = 0

\frac{1}{x} \frac{d y}{d x} = \frac{1}{x ^{2} + 1}

f (x) = 5 cos (x)

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