What is the integral of (-9x+27)/(2x-3) ?

The integral of (-9x+27)/(2x-3) is 1/4 (-9(2x-3)+27ln|2x-3|)+C

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integral of (-9x+27)/(2x-3)

\int \frac{- 9 x + 27}{2 x - 3} d x

\frac{1}{4} (- 9 (2 x - 3) + 27 ln ∣ 2 x - 3 ∣) + C

Solution steps

\int \frac{- 9 x + 27}{2 x - 3} d x

Apply u-substitution

: \int \frac{- 9 u + 27}{4 u} d u

= \int \frac{- 9 u + 27}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{- 9 u + 27}{u} d u

Expand

\frac{- 9 u + 27}{u} : - 9 + \frac{27}{u}

= \frac{1}{4} \cdot \int - 9 + \frac{27}{u} d u

Apply the Sum Rule:

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

= \frac{1}{4} (- \int 9 d u + \int \frac{27}{u} d u)

\int 9 d u = 9 u

\int \frac{27}{u} d u = 27 ln ∣ u ∣

= \frac{1}{4} (- 9 u + 27 ln ∣ u ∣)

Substitute back

u = 2 x - 3

= \frac{1}{4} (- 9 (2 x - 3) + 27 ln ∣ 2 x - 3 ∣)

Add a constant to the solution

= \frac{1}{4} (- 9 (2 x - 3) + 27 ln ∣ 2 x - 3 ∣) + C

\int (x + 1) y^{2} sin (\frac{x y}{3}) d x

n = 1 \sum \infty \frac{n - 5}{n + 5}

\frac{d}{d x} (x^{2} + y^{2} - 5)

\frac{4 x}{x - 5}

\frac{d x}{d r} + e^{r - x} = 0

What is the integral of (-9x+27)/(2x-3) ?
The integral of (-9x+27)/(2x-3) is 1/4 (-9(2x-3)+27ln|2x-3|)+C