What is the integral of (2x+1)/(3x^2+3x+4) ?

The integral of (2x+1)/(3x^2+3x+4) is 1/3 ln|12x^2+12x+16|+C

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integral of (2x+1)/(3x^2+3x+4)

\int \frac{2 x + 1}{3 x ^{2} + 3 x + 4} d x

\frac{1}{3} ln 12 x^{2} + 12 x + 16 + C

Solution steps

\int \frac{2 x + 1}{3 x ^{2} + 3 x + 4} d x

Expand

\frac{2 x + 1}{3 x ^{2} + 3 x + 4} : \frac{2 x}{3 x ^{2} + 3 x + 4} + \frac{1}{3 x ^{2} + 3 x + 4}

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}{3 x ^{2} + 3 x + 4} d x + \int \frac{1}{3 x ^{2} + 3 x + 4} d x

\int \frac{2 x}{3 x ^{2} + 3 x + 4} d x = 4 (\frac{1}{12} ln 12 x^{2} + 12 x + 16 - \frac{1}{2 39} arctan (\frac{2 3}{13} x + \frac{3}{13}))

\int \frac{1}{3 x ^{2} + 3 x + 4} d x = \frac{2}{39} arctan (\frac{2 3}{13} x + \frac{3}{13})

= 4 (\frac{1}{12} ln 12 x^{2} + 12 x + 16 - \frac{1}{2 39} arctan (\frac{2 3}{13} x + \frac{3}{13})) + \frac{2}{39} arctan (\frac{2 3}{13} x + \frac{3}{13})

Simplify

4 (\frac{1}{12} ln 12 x^{2} + 12 x + 16 - \frac{1}{2 39} arctan (\frac{2 3}{13} x + \frac{3}{13})) + \frac{2}{39} arctan (\frac{2 3}{13} x + \frac{3}{13}) : \frac{1}{3} ln 12 x^{2} + 12 x + 16

= \frac{1}{3} ln 12 x^{2} + 12 x + 16

Add a constant to the solution

= \frac{1}{3} ln 12 x^{2} + 12 x + 16 + C

3 x + 3, 4 x + 4, - 1, 3

\frac{\partial}{\partial x} (ln (2 x y + 4))

\frac{\partial}{\partial y} (arctan (x + 6 y))

y = x^{2} + 3

\int_{1}^{4} \frac{7}{x ^{2}} d x

What is the integral of (2x+1)/(3x^2+3x+4) ?
The integral of (2x+1)/(3x^2+3x+4) is 1/3 ln|12x^2+12x+16|+C