What is the integral of (x-3)/(x^2+9) ?

The integral of (x-3)/(x^2+9) is 1/2 ln|x^2+9|-arctan(x/3)+C

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integral of (x-3)/(x^2+9)

\int \frac{x - 3}{x ^{2} + 9} d x

\frac{1}{2} ln x^{2} + 9 - arctan (\frac{x}{3}) + C

Solution steps

\int \frac{x - 3}{x ^{2} + 9} d x

Expand

\frac{x - 3}{x ^{2} + 9} : \frac{x}{x ^{2} + 9} - \frac{3}{x ^{2} + 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{x ^{2} + 9} d x - \int \frac{3}{x ^{2} + 9} d x

\int \frac{x}{x ^{2} + 9} d x = \frac{1}{2} ln x^{2} + 9

\int \frac{3}{x ^{2} + 9} d x = arctan (\frac{x}{3})

= \frac{1}{2} ln x^{2} + 9 - arctan (\frac{x}{3})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 9 - arctan (\frac{x}{3}) + C

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What is the integral of (x-3)/(x^2+9) ?
The integral of (x-3)/(x^2+9) is 1/2 ln|x^2+9|-arctan(x/3)+C