What is the integral of x/(x^2-20x) ?

The integral of x/(x^2-20x) is ln|x-20|+C

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integral of x/(x^2-20x)

\int \frac{x}{x ^{2} - 20 x} d x

ln ∣ x - 20 ∣ + C

Solution steps

\int \frac{x}{x ^{2} - 20 x} d x

Factor

\frac{x}{x ^{2} - 20 x}

= \int \frac{x}{x ( x - 20 )} d x

Cancel the common factor:

x

= \int \frac{1}{x - 20} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x - 20

= ln ∣ x - 20 ∣

Add a constant to the solution

= ln ∣ x - 20 ∣ + C

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