What is the integral of 1/(x^2-x^3) ?

The integral of 1/(x^2-x^3) is ln|x|-1/x-ln|x-1|+C

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integral of 1/(x^2-x^3)

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{2} - x ^{3}} : - \frac{- x - 1}{x ^{2}} - \frac{1}{x - 1}

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

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

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

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

Simplify

= 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 x} (ln (2 x - y))

\int_{0}^{9} 9 + 3 x d x

3 t^{2}

y^{^{''}} - 4 y^{^{'}} + 53 y = 5 e^{2 t} cos (7 t)

cos^{10} (sin (7 x))

What is the integral of 1/(x^2-x^3) ?
The integral of 1/(x^2-x^3) is ln|x|-1/x-ln|x-1|+C