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

The integral of (3x+2)/(9x^2-12x+8) is 1/6 ln|9x^2-12x+8|+2/3 arctan(3/2 x-1)+C

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

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

\frac{1}{6} ln 9 x^{2} - 12 x + 8 + \frac{2}{3} arctan (\frac{3}{2} x - 1) + C

Solution steps

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

Expand

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

\int \frac{3 x}{9 x ^{2} - 12 x + 8} d x = \frac{1}{6} ln 9 x^{2} - 12 x + 8 + \frac{1}{3} arctan (\frac{3}{2} x - 1)

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

= \frac{1}{6} ln 9 x^{2} - 12 x + 8 + \frac{1}{3} arctan (\frac{3}{2} x - 1) + \frac{1}{3} arctan (\frac{3}{2} x - 1)

Add similar elements:

\frac{1}{3} arctan (\frac{3}{2} x - 1) + \frac{1}{3} arctan (\frac{3}{2} x - 1) = \frac{2}{3} arctan (\frac{3}{2} x - 1)

= \frac{1}{6} ln 9 x^{2} - 12 x + 8 + \frac{2}{3} arctan (\frac{3}{2} x - 1)

Add a constant to the solution

= \frac{1}{6} ln 9 x^{2} - 12 x + 8 + \frac{2}{3} arctan (\frac{3}{2} x - 1) + C

y = x^{e^{x}}

\int x sin (\frac{1}{9} x) d x

\int_{- 4}^{4} x (16 - x^{2}) d x

x \to 6 lim (5 x + 6)

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

What is the integral of (3x+2)/(9x^2-12x+8) ?
The integral of (3x+2)/(9x^2-12x+8) is 1/6 ln|9x^2-12x+8|+2/3 arctan(3/2 x-1)+C