What is the integral of (3x-2)/(1-6x-9x^2) ?

The integral of (3x-2)/(1-6x-9x^2) is

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integral of (3x-2)/(1-6x-9x^2)

\int \frac{3 x - 2}{1 - 6 x - 9 x ^{2}} d x

\frac{1}{6 2} (3 x - 2) (ln \frac{1}{2} (3 x + 1) + 1 - ln \frac{1}{2} (3 x + 1) - 1) + \frac{1}{2 2} x ln \frac{3 x + 1}{2} - 1 - \frac{1}{2 2} x ln \frac{3 x + 1}{2} + 1 + \frac{- 2 - 1}{6 2} ln \frac{1}{2} (3 x + 1) + 1 + \frac{- 2 + 1}{6 2} ln \frac{1}{2} (3 x + 1) - 1 + \frac{1}{3} + C

Solution steps

\int \frac{3 x - 2}{1 - 6 x - 9 x ^{2}} d x

Apply u-substitution

: \int \frac{u}{3 ( - u ^{2} - 6 u - 7 )} d u

= \int \frac{u}{3 ( - u ^{2} - 6 u - 7 )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{u}{- u ^{2} - 6 u - 7} d u

Apply Integration By Parts

: \frac{1}{2 2} u (ln \frac{1}{2} (u + 3) + 1 - ln \frac{1}{2} (u + 3) - 1) - \int \frac{1}{2 2} (ln \frac{1}{2} (u + 3) + 1 - ln \frac{1}{2} (u + 3) - 1) d u

= \frac{1}{3} (\frac{1}{2 2} u (ln \frac{1}{2} (u + 3) + 1 - ln \frac{1}{2} (u + 3) - 1) - \int \frac{1}{2 2} (ln \frac{1}{2} (u + 3) + 1 - ln \frac{1}{2} (u + 3) - 1) d u)

\int \frac{1}{2 2} (ln \frac{1}{2} (u + 3) + 1 - ln \frac{1}{2} (u + 3) - 1) d u = \frac{1}{2 2} (- u ln \frac{u + 3}{2} - 1 + ln \frac{1}{2} (u + 3) + 1 (u + 3) + 2 ln \frac{1}{2} (u + 3) + 1 - (3 - 2) ln \frac{u + 3}{2} - 1 - 22)

= \frac{1}{3} (\frac{1}{2 2} u (ln \frac{1}{2} (u + 3) + 1 - ln \frac{1}{2} (u + 3) - 1) - \frac{1}{2 2} (- u ln \frac{u + 3}{2} - 1 + ln \frac{1}{2} (u + 3) + 1 (u + 3) + 2 ln \frac{1}{2} (u + 3) + 1 - (3 - 2) ln \frac{u + 3}{2} - 1 - 22))

Substitute back

u = 3 x - 2

= \frac{1}{3} (\frac{1}{2 2} (3 x - 2) (ln \frac{1}{2} (3 x - 2 + 3) + 1 - ln \frac{1}{2} (3 x - 2 + 3) - 1) - \frac{1}{2 2} (- (3 x - 2) ln \frac{3 x - 2 + 3}{2} - 1 + ln \frac{1}{2} (3 x - 2 + 3) + 1 (3 x - 2 + 3) + 2 ln \frac{1}{2} (3 x - 2 + 3) + 1 - (3 - 2) ln \frac{3 x - 2 + 3}{2} - 1 - 22))

Simplify

\frac{1}{3} (\frac{1}{2 2} (3 x - 2) (ln \frac{1}{2} (3 x - 2 + 3) + 1 - ln \frac{1}{2} (3 x - 2 + 3) - 1) - \frac{1}{2 2} (- (3 x - 2) ln \frac{3 x - 2 + 3}{2} - 1 + ln \frac{1}{2} (3 x - 2 + 3) + 1 (3 x - 2 + 3) + 2 ln \frac{1}{2} (3 x - 2 + 3) + 1 - (3 - 2) ln \frac{3 x - 2 + 3}{2} - 1 - 22)) : \frac{1}{6 2} (3 x - 2) (ln \frac{1}{2} (3 x + 1) + 1 - ln \frac{1}{2} (3 x + 1) - 1) + \frac{1}{2 2} x ln \frac{3 x + 1}{2} - 1 - \frac{1}{2 2} x ln \frac{3 x + 1}{2} + 1 + \frac{- 2 - 1}{6 2} ln \frac{1}{2} (3 x + 1) + 1 + \frac{- 2 + 1}{6 2} ln \frac{1}{2} (3 x + 1) - 1 + \frac{1}{3}

= \frac{1}{6 2} (3 x - 2) (ln \frac{1}{2} (3 x + 1) + 1 - ln \frac{1}{2} (3 x + 1) - 1) + \frac{1}{2 2} x ln \frac{3 x + 1}{2} - 1 - \frac{1}{2 2} x ln \frac{3 x + 1}{2} + 1 + \frac{- 2 - 1}{6 2} ln \frac{1}{2} (3 x + 1) + 1 + \frac{- 2 + 1}{6 2} ln \frac{1}{2} (3 x + 1) - 1 + \frac{1}{3}

Add a constant to the solution

= \frac{1}{6 2} (3 x - 2) (ln \frac{1}{2} (3 x + 1) + 1 - ln \frac{1}{2} (3 x + 1) - 1) + \frac{1}{2 2} x ln \frac{3 x + 1}{2} - 1 - \frac{1}{2 2} x ln \frac{3 x + 1}{2} + 1 + \frac{- 2 - 1}{6 2} ln \frac{1}{2} (3 x + 1) + 1 + \frac{- 2 + 1}{6 2} ln \frac{1}{2} (3 x + 1) - 1 + \frac{1}{3} + C

\int \frac{6 x - 5 x + 1}{2 x} d x

\frac{d}{d x} ((4 x + 5)^{3})

\int_{0}^{1} (\frac{x - x}{3}) d x

cot (x^{6})

\frac{1}{y} \frac{d y}{d x} = e^{x} + lo g_{e} (y)

What is the integral of (3x-2)/(1-6x-9x^2) ?
The integral of (3x-2)/(1-6x-9x^2) is