What is the integral of (x^2+5x)/((x-3)(x^2+9)) ?

The integral of (x^2+5x)/((x-3)(x^2+9)) is 4/3 ln|x-3|+1/3 (-1/2 ln|(x^2)/9+1|+4arctan(x/3))+C

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integral of (x^2+5x)/((x-3)(x^2+9))

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

\frac{4}{3} ln ∣ x - 3 ∣ + \frac{1}{3} (- \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 4 arctan (\frac{x}{3})) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 5 x}{( x - 3 ) ( x ^{2} + 9 )} : \frac{4}{3 ( x - 3 )} + \frac{- x + 12}{3 ( x ^{2} + 9 )}

= \int \frac{4}{3 ( x - 3 )} + \frac{- x + 12}{3 ( 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{4}{3 ( x - 3 )} d x + \int \frac{- x + 12}{3 ( x ^{2} + 9 )} d x

\int \frac{4}{3 ( x - 3 )} d x = \frac{4}{3} ln ∣ x - 3 ∣

\int \frac{- x + 12}{3 ( x ^{2} + 9 )} d x = \frac{1}{3} (- \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 4 arctan (\frac{x}{3}))

= \frac{4}{3} ln ∣ x - 3 ∣ + \frac{1}{3} (- \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 4 arctan (\frac{x}{3}))

Add a constant to the solution

= \frac{4}{3} ln ∣ x - 3 ∣ + \frac{1}{3} (- \frac{1}{2} ln \frac{x ^{2}}{9} + 1 + 4 arctan (\frac{x}{3})) + C

y = x^{2} - 37, y = 11 - 2 x

y = e^{- 6 x^{5}}

\int \frac{1}{x ( 1 + ( ln ( x ) ) ^{2} )} d x

- 2 (x - 8)^{2}

\int \frac{1}{x x ^{2} + 7} d x

What is the integral of (x^2+5x)/((x-3)(x^2+9)) ?
The integral of (x^2+5x)/((x-3)(x^2+9)) is 4/3 ln|x-3|+1/3 (-1/2 ln|(x^2)/9+1|+4arctan(x/3))+C