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

The integral of ((x+1))/(2x^2+6x+9) is 1/4 ln|4x^2+12x+18|-1/6 arctan(2/3 x+1)+C

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

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

\frac{1}{4} ln 4 x^{2} + 12 x + 18 - \frac{1}{6} arctan (\frac{2}{3} x + 1) + C

Solution steps

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

Expand

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

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

\int \frac{x}{2 x ^{2} + 6 x + 9} d x = \frac{1}{4} ln 4 x^{2} + 12 x + 18 - \frac{1}{2} arctan (\frac{2}{3} x + 1)

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

= \frac{1}{4} ln 4 x^{2} + 12 x + 18 - \frac{1}{2} arctan (\frac{2}{3} x + 1) + \frac{1}{3} arctan (\frac{2}{3} x + 1)

Add similar elements:

- \frac{1}{2} arctan (\frac{2}{3} x + 1) + \frac{1}{3} arctan (\frac{2}{3} x + 1) = - \frac{1}{6} arctan (\frac{2}{3} x + 1)

= \frac{1}{4} ln 4 x^{2} + 12 x + 18 - \frac{1}{6} arctan (\frac{2}{3} x + 1)

Add a constant to the solution

= \frac{1}{4} ln 4 x^{2} + 12 x + 18 - \frac{1}{6} arctan (\frac{2}{3} x + 1) + C

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What is the integral of ((x+1))/(2x^2+6x+9) ?
The integral of ((x+1))/(2x^2+6x+9) is 1/4 ln|4x^2+12x+18|-1/6 arctan(2/3 x+1)+C