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

Take the constant out:

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

= - \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