What is the integral of (x-3)/((x+1)(x^2+9)) ?

The integral of (x-3)/((x+1)(x^2+9)) is -2/5 ln|x+1|+1/5 (ln|x^2+9|+arctan(x/3))+C

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integral of (x-3)/((x+1)(x^2+9))

\int \frac{x - 3}{( x + 1 ) ( x ^{2} + 9 )} d x

- \frac{2}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (ln x^{2} + 9 + arctan (\frac{x}{3})) + C

Solution steps

\int \frac{x - 3}{( x + 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{x - 3}{( x + 1 ) ( x ^{2} + 9 )} : - \frac{2}{5 ( x + 1 )} + \frac{2 x + 3}{5 ( x ^{2} + 9 )}

= \int - \frac{2}{5 ( x + 1 )} + \frac{2 x + 3}{5 ( x ^{2} + 9 )} d x

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{2}{5 ( x + 1 )} d x + \int \frac{2 x + 3}{5 ( x ^{2} + 9 )} d x

\int \frac{2}{5 ( x + 1 )} d x = \frac{2}{5} ln ∣ x + 1 ∣

\int \frac{2 x + 3}{5 ( x ^{2} + 9 )} d x = \frac{1}{5} (ln x^{2} + 9 + arctan (\frac{x}{3}))

= - \frac{2}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (ln x^{2} + 9 + arctan (\frac{x}{3}))

Add a constant to the solution

= - \frac{2}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (ln x^{2} + 9 + arctan (\frac{x}{3})) + C

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What is the integral of (x-3)/((x+1)(x^2+9)) ?
The integral of (x-3)/((x+1)(x^2+9)) is -2/5 ln|x+1|+1/5 (ln|x^2+9|+arctan(x/3))+C