What is the integral of (5x+1)/(9x^2+6x+3) ?

The integral of (5x+1)/(9x^2+6x+3) is 5/18 ln(3|3x^2+2x+1|)-(sqrt(2))/9 arctan(1/(sqrt(2))(3x+1))+C

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integral of (5x+1)/(9x^2+6x+3)

\int \frac{5 x + 1}{9 x ^{2} + 6 x + 3} d x

\frac{5}{18} ln (3 3 x^{2} + 2 x + 1) - \frac{2}{9} arctan (\frac{1}{2} (3 x + 1)) + C

Solution steps

\int \frac{5 x + 1}{9 x ^{2} + 6 x + 3} d x

Expand

\frac{5 x + 1}{9 x ^{2} + 6 x + 3} : \frac{5 x}{9 x ^{2} + 6 x + 3} + \frac{1}{9 x ^{2} + 6 x + 3}

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{5 x}{9 x ^{2} + 6 x + 3} d x + \int \frac{1}{9 x ^{2} + 6 x + 3} d x

\int \frac{5 x}{9 x ^{2} + 6 x + 3} d x = \frac{5}{18} (ln (3 3 x^{2} + 2 x + 1) - 2 arctan (\frac{3 x + 1}{2}))

\int \frac{1}{9 x ^{2} + 6 x + 3} d x = \frac{1}{3 2} arctan (\frac{3 x + 1}{2})

= \frac{5}{18} (ln (3 3 x^{2} + 2 x + 1) - 2 arctan (\frac{3 x + 1}{2})) + \frac{1}{3 2} arctan (\frac{3 x + 1}{2})

Simplify

\frac{5}{18} (ln (3 3 x^{2} + 2 x + 1) - 2 arctan (\frac{3 x + 1}{2})) + \frac{1}{3 2} arctan (\frac{3 x + 1}{2}) : \frac{5}{18} ln (3 3 x^{2} + 2 x + 1) - \frac{2}{9} arctan (\frac{1}{2} (3 x + 1))

= \frac{5}{18} ln (3 3 x^{2} + 2 x + 1) - \frac{2}{9} arctan (\frac{1}{2} (3 x + 1))

Add a constant to the solution

= \frac{5}{18} ln (3 3 x^{2} + 2 x + 1) - \frac{2}{9} arctan (\frac{1}{2} (3 x + 1)) + C

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What is the integral of (5x+1)/(9x^2+6x+3) ?
The integral of (5x+1)/(9x^2+6x+3) is 5/18 ln(3|3x^2+2x+1|)-(sqrt(2))/9 arctan(1/(sqrt(2))(3x+1))+C