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

The integral of (2x+1)/(x^2+x+3) is ln|4x^2+4x+12|+C

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

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

ln 4 x^{2} + 4 x + 12 + C

Solution steps

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

Expand

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

\int \frac{2 x}{x ^{2} + x + 3} d x = ln 4 x^{2} + 4 x + 12 - \frac{2}{11} arctan (\frac{2 x + 1}{11})

\int \frac{1}{x ^{2} + x + 3} d x = \frac{2}{11} arctan (\frac{2 x + 1}{11})

= ln 4 x^{2} + 4 x + 12 - \frac{2}{11} arctan (\frac{2 x + 1}{11}) + \frac{2}{11} arctan (\frac{2 x + 1}{11})

Add similar elements:

- \frac{2}{11} arctan (\frac{2 x + 1}{11}) + \frac{2}{11} arctan (\frac{2 x + 1}{11}) = 0

= ln 4 x^{2} + 4 x + 12

Add a constant to the solution

= ln 4 x^{2} + 4 x + 12 + C

f (x) = \frac{x ^{2} - 81}{x - 9}

\int sin (12 x) cos (3 x) d x

\frac{d}{d x} (\frac{x ^{2}}{( 1 - y ^{2} )})

\int (\frac{y ^{2}}{4})^{2} d y

\frac{d}{d x} (e^{x^{3} + 6 x^{2}})

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