What is the integral of 1/((x+3)(9x^2+4)) ?

The integral of 1/((x+3)(9x^2+4)) is 1/85 ln|x+3|+1/85 (-1/2 ln|9x^2+4|+9/2 arctan(3/2 x))+C

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integral of 1/((x+3)(9x^2+4))

\int \frac{1}{( x + 3 ) ( 9 x ^{2} + 4 )} d x

\frac{1}{85} ln ∣ x + 3 ∣ + \frac{1}{85} (- \frac{1}{2} ln 9 x^{2} + 4 + \frac{9}{2} arctan (\frac{3}{2} x)) + C

Solution steps

\int \frac{1}{( x + 3 ) ( 9 x ^{2} + 4 )} d x

Take the partial fraction of

\frac{1}{( x + 3 ) ( 9 x ^{2} + 4 )} : \frac{1}{85 ( x + 3 )} + \frac{- 9 x + 27}{85 ( 9 x ^{2} + 4 )}

= \int \frac{1}{85 ( x + 3 )} + \frac{- 9 x + 27}{85 ( 9 x ^{2} + 4 )} 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}{85 ( x + 3 )} d x + \int \frac{- 9 x + 27}{85 ( 9 x ^{2} + 4 )} d x

\int \frac{1}{85 ( x + 3 )} d x = \frac{1}{85} ln ∣ x + 3 ∣

\int \frac{- 9 x + 27}{85 ( 9 x ^{2} + 4 )} d x = \frac{1}{85} (- \frac{1}{2} ln 9 x^{2} + 4 + \frac{9}{2} arctan (\frac{3}{2} x))

= \frac{1}{85} ln ∣ x + 3 ∣ + \frac{1}{85} (- \frac{1}{2} ln 9 x^{2} + 4 + \frac{9}{2} arctan (\frac{3}{2} x))

Add a constant to the solution

= \frac{1}{85} ln ∣ x + 3 ∣ + \frac{1}{85} (- \frac{1}{2} ln 9 x^{2} + 4 + \frac{9}{2} arctan (\frac{3}{2} x)) + C

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What is the integral of 1/((x+3)(9x^2+4)) ?
The integral of 1/((x+3)(9x^2+4)) is 1/85 ln|x+3|+1/85 (-1/2 ln|9x^2+4|+9/2 arctan(3/2 x))+C