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

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

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

\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})) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Complete the square

9 x^{2} - 12 x + 8 : 9 (x - \frac{2}{3})^{2} + 4

= 2 \cdot \int \frac{x}{9 ( x - \frac{2}{3} ) ^{2} + 4} d x

Apply u-substitution

= 2 \cdot \int \frac{3 u + 2}{3 ( 9 u ^{2} + 4 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{3} \cdot \int \frac{3 u + 2}{9 u ^{2} + 4} d u

Expand

\frac{3 u + 2}{9 u ^{2} + 4} : \frac{3 u}{9 u ^{2} + 4} + \frac{2}{9 u ^{2} + 4}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{3} (\int \frac{3 u}{9 u ^{2} + 4} d u + \int \frac{2}{9 u ^{2} + 4} d u)

\int \frac{3 u}{9 u ^{2} + 4} d u = \frac{1}{6} ln 9 u^{2} + 4

\int \frac{2}{9 u ^{2} + 4} d u = \frac{1}{3} arctan (\frac{3}{2} u)

= 2 \cdot \frac{1}{3} (\frac{1}{6} ln 9 u^{2} + 4 + \frac{1}{3} arctan (\frac{3}{2} u))

Substitute back

u = x - \frac{2}{3}

= 2 \cdot \frac{1}{3} (\frac{1}{6} ln 9 (x - \frac{2}{3})^{2} + 4 + \frac{1}{3} arctan (\frac{3}{2} (x - \frac{2}{3})))

Simplify

2 \cdot \frac{1}{3} (\frac{1}{6} ln 9 (x - \frac{2}{3})^{2} + 4 + \frac{1}{3} arctan (\frac{3}{2} (x - \frac{2}{3}))) : \frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2}))

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

Add a constant to the solution

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

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