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

The integral of x/((x^2-9)) is 1/2 ln|x^2-9|+C

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

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

\frac{1}{2} ln x^{2} - 9 + C

Solution steps

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

Apply u-substitution

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 9

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

Add a constant to the solution

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

What is the integral of x/((x^2-9)) ?
The integral of x/((x^2-9)) is 1/2 ln|x^2-9|+C