What is the integral of (2x+3)/(9x^2-12x+8) ?

The integral of (2x+3)/(9x^2-12x+8) is 1/9 (ln|9x^2-12x+8|+2arctan((3x-2)/2))+1/2 arctan(3/2 x-1)+C

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integral of (2x+3)/(9x^2-12x+8)

\int \frac{2 x + 3}{9 x ^{2} - 12 x + 8} d x

\frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2})) + \frac{1}{2} arctan (\frac{3}{2} x - 1) + C

Solution steps

\int \frac{2 x + 3}{9 x ^{2} - 12 x + 8} d x

Expand

\frac{2 x + 3}{9 x ^{2} - 12 x + 8} : \frac{2 x}{9 x ^{2} - 12 x + 8} + \frac{3}{9 x ^{2} - 12 x + 8}

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}{9 x ^{2} - 12 x + 8} d x + \int \frac{3}{9 x ^{2} - 12 x + 8} d x

\int \frac{2 x}{9 x ^{2} - 12 x + 8} d x = \frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2}))

\int \frac{3}{9 x ^{2} - 12 x + 8} d x = \frac{1}{2} arctan (\frac{3}{2} x - 1)

= \frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2})) + \frac{1}{2} arctan (\frac{3}{2} x - 1)

Add a constant to the solution

= \frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2})) + \frac{1}{2} arctan (\frac{3}{2} x - 1) + C

x \to \infty lim (\frac{1 - 5 e ^{2 x}}{3 e ^{x} + 4 e ^{2 x}})

\frac{d}{d x} (6 tan (x))

\frac{d}{d x} (ln (csc (x) + cot (x)))

x \to \infty lim (- 2 x + 8)

\frac{\partial}{\partial x} (\frac{x}{x ^{5} - y ^{4}})

What is the integral of (2x+3)/(9x^2-12x+8) ?
The integral of (2x+3)/(9x^2-12x+8) is 1/9 (ln|9x^2-12x+8|+2arctan((3x-2)/2))+1/2 arctan(3/2 x-1)+C