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

The integral of (x+1)/(x^2+x+2) is 1/2 ln|4x^2+4x+8|+1/(sqrt(7))arctan(1/(sqrt(7))(2x+1))+C

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

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

\frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (2 x + 1)) + C

Solution steps

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

Expand

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

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

\int \frac{x}{x ^{2} + x + 2} d x = \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7})

\int \frac{1}{x ^{2} + x + 2} d x = \frac{2}{7} arctan (\frac{2 x + 1}{7})

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7}) + \frac{2}{7} arctan (\frac{2 x + 1}{7})

Simplify

\frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7}) + \frac{2}{7} arctan (\frac{2 x + 1}{7}) : \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (2 x + 1))

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (2 x + 1))

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (2 x + 1)) + C

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What is the integral of (x+1)/(x^2+x+2) ?
The integral of (x+1)/(x^2+x+2) is 1/2 ln|4x^2+4x+8|+1/(sqrt(7))arctan(1/(sqrt(7))(2x+1))+C