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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{1}{3 x ^{2}} d x = - \frac{1}{3 x}

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

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

Add a constant to the solution

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} (- \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(x^2+9)) ?
The integral of (x+3)/(x^2(x^2+9)) is 1/9 ln|x|-1/(3x)+1/9 (-1/2 ln|x^2+9|-arctan(x/3))+C