What is the integral of ((3x+2)(x-4))/(x-3) ?

The integral of ((3x+2)(x-4))/(x-3) is 1/6 (9x^2-6x-66ln|3x-9|-8)+C

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integral of ((3x+2)(x-4))/(x-3)

\int \frac{( 3 x + 2 ) ( x - 4 )}{x - 3} d x

\frac{1}{6} (9 x^{2} - 6 x - 66 ln ∣ 3 x - 9 ∣ - 8) + C

Solution steps

\int \frac{( 3 x + 2 ) ( x - 4 )}{x - 3} d x

Apply u-substitution

= \int \frac{u ( u - 14 )}{3 ( u - 11 )} 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 - 14 )}{u - 11} d u

Long division

\frac{u ( u - 14 )}{u - 11} : u - 3 - \frac{33}{u - 11}

= \frac{1}{3} \cdot \int u - 3 - \frac{33}{u - 11} 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}{3} (\int u d u - \int 3 d u - \int \frac{33}{u - 11} d u)

\int u d u = \frac{u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{33}{u - 11} d u = 33 ln ∣ u - 11 ∣

= \frac{1}{3} (\frac{u ^{2}}{2} - 3 u - 33 ln ∣ u - 11 ∣)

Substitute back

u = 3 x + 2

= \frac{1}{3} (\frac{( 3 x + 2 ) ^{2}}{2} - 3 (3 x + 2) - 33 ln ∣ 3 x + 2 - 11 ∣)

Simplify

\frac{1}{3} (\frac{( 3 x + 2 ) ^{2}}{2} - 3 (3 x + 2) - 33 ln ∣ 3 x + 2 - 11 ∣) : \frac{1}{6} (9 x^{2} - 6 x - 66 ln ∣ 3 x - 9 ∣ - 8)

= \frac{1}{6} (9 x^{2} - 6 x - 66 ln ∣ 3 x - 9 ∣ - 8)

Add a constant to the solution

= \frac{1}{6} (9 x^{2} - 6 x - 66 ln ∣ 3 x - 9 ∣ - 8) + C

What is the integral of ((3x+2)(x-4))/(x-3) ?
The integral of ((3x+2)(x-4))/(x-3) is 1/6 (9x^2-6x-66ln|3x-9|-8)+C