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

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

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

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

ln x^{2} - 9 + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 9

= 2 \cdot \frac{1}{2} ln x^{2} - 9

Simplify

2 \cdot \frac{1}{2} ln x^{2} - 9 : ln x^{2} - 9

= ln x^{2} - 9

Add a constant to the solution

= ln x^{2} - 9 + C

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