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

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

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integral of x/(x-x^2)

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

- ln ∣ 1 - x ∣ + C

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 1 - x

= - ln ∣ 1 - x ∣

Add a constant to the solution

= - ln ∣ 1 - x ∣ + C

k = 2 \sum \infty (\frac{1}{3})^{k}

y = arcsin (x) + 1 - x^{2}

\frac{\partial}{\partial v} (uv - 1)

\frac{d}{d x} (4 x + \frac{64}{x})

f (x) = 3 x - 9