What is the integral of (2x)/(x^2-9x) ?

The integral of (2x)/(x^2-9x) is 2ln|x-9|+C

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integral of (2x)/(x^2-9x)

\int \frac{2 x}{x ^{2} - 9 x} d x

2 ln ∣ x - 9 ∣ + C

Solution steps

\int \frac{2 x}{x ^{2} - 9 x} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{x ^{2} - 9 x} d x

Factor

\frac{x}{x ^{2} - 9 x}

= 2 \cdot \int \frac{x}{x ( x - 9 )} d x

Cancel the common factor:

x

= 2 \cdot \int \frac{1}{x - 9} d x

Apply u-substitution

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

= 2 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

u = x - 9

= 2 ln ∣ x - 9 ∣

Add a constant to the solution

= 2 ln ∣ x - 9 ∣ + C

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What is the integral of (2x)/(x^2-9x) ?
The integral of (2x)/(x^2-9x) is 2ln|x-9|+C