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

The integral of (x+1)/(x^2+x+6) is 1/2 ln|4x^2+4x+24|+1/(sqrt(23))arctan(1/(sqrt(23))(2x+1))+C

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

integral

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

\frac{1}{2} ln 4 x^{2} + 4 x + 24 + \frac{1}{23} arctan (\frac{1}{23} (2 x + 1)) + C

Solution steps

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

Expand

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

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

\int \frac{x}{x ^{2} + x + 6} d x = \frac{1}{2} ln 4 x^{2} + 4 x + 24 - \frac{1}{23} arctan (\frac{2 x + 1}{23})

\int \frac{1}{x ^{2} + x + 6} d x = \frac{2}{23} arctan (\frac{2 x + 1}{23})

= \frac{1}{2} ln 4 x^{2} + 4 x + 24 - \frac{1}{23} arctan (\frac{2 x + 1}{23}) + \frac{2}{23} arctan (\frac{2 x + 1}{23})

Simplify

\frac{1}{2} ln 4 x^{2} + 4 x + 24 - \frac{1}{23} arctan (\frac{2 x + 1}{23}) + \frac{2}{23} arctan (\frac{2 x + 1}{23}) : \frac{1}{2} ln 4 x^{2} + 4 x + 24 + \frac{1}{23} arctan (\frac{1}{23} (2 x + 1))

= \frac{1}{2} ln 4 x^{2} + 4 x + 24 + \frac{1}{23} arctan (\frac{1}{23} (2 x + 1))

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} + 4 x + 24 + \frac{1}{23} arctan (\frac{1}{23} (2 x + 1)) + C

3 e^{3} x

ln (1 - z)

\frac{2 x}{( 1 + x ^{2} )}

\frac{1}{3 ^{x}}

\frac{e ^{x}}{( 2 + e ^{x} )}

What is the integral of (x+1)/(x^2+x+6) ?
The integral of (x+1)/(x^2+x+6) is 1/2 ln|4x^2+4x+24|+1/(sqrt(23))arctan(1/(sqrt(23))(2x+1))+C