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

The integral of (x+1)/(2x^2+2x+3) is 1/4 ln|4x^2+4x+6|+1/(2sqrt(5))arctan(1/(sqrt(5))(2x+1))+C

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

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

\frac{1}{4} ln 4 x^{2} + 4 x + 6 + \frac{1}{2 5} arctan (\frac{1}{5} (2 x + 1)) + C

Solution steps

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

Expand

\frac{x + 1}{2 x ^{2} + 2 x + 3} : \frac{x}{2 x ^{2} + 2 x + 3} + \frac{1}{2 x ^{2} + 2 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{x}{2 x ^{2} + 2 x + 3} d x + \int \frac{1}{2 x ^{2} + 2 x + 3} d x

\int \frac{x}{2 x ^{2} + 2 x + 3} d x = \frac{1}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{2 x + 1}{5})

\int \frac{1}{2 x ^{2} + 2 x + 3} d x = \frac{1}{5} arctan (\frac{2 x + 1}{5})

= \frac{1}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{2 x + 1}{5}) + \frac{1}{5} arctan (\frac{2 x + 1}{5})

Simplify

\frac{1}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{2 x + 1}{5}) + \frac{1}{5} arctan (\frac{2 x + 1}{5}) : \frac{1}{4} ln 4 x^{2} + 4 x + 6 + \frac{1}{2 5} arctan (\frac{1}{5} (2 x + 1))

= \frac{1}{4} ln 4 x^{2} + 4 x + 6 + \frac{1}{2 5} arctan (\frac{1}{5} (2 x + 1))

Add a constant to the solution

= \frac{1}{4} ln 4 x^{2} + 4 x + 6 + \frac{1}{2 5} arctan (\frac{1}{5} (2 x + 1)) + C

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What is the integral of (x+1)/(2x^2+2x+3) ?
The integral of (x+1)/(2x^2+2x+3) is 1/4 ln|4x^2+4x+6|+1/(2sqrt(5))arctan(1/(sqrt(5))(2x+1))+C