What is the integral of (4x^2+4x-6)/((x-6)(x-3)^2) ?

The integral of (4x^2+4x-6)/((x-6)(x-3)^2) is 18ln|x-6|-14ln|x-3|+(14)/(x-3)+C

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integral of (4x^2+4x-6)/((x-6)(x-3)^2)

\int \frac{4 x ^{2} + 4 x - 6}{( x - 6 ) ( x - 3 ) ^{2}} d x

18 ln ∣ x - 6 ∣ - 14 ln ∣ x - 3 ∣ + \frac{14}{x - 3} + C

Solution steps

\int \frac{4 x ^{2} + 4 x - 6}{( x - 6 ) ( x - 3 ) ^{2}} d x

Take the partial fraction of

\frac{4 x ^{2} + 4 x - 6}{( x - 6 ) ( x - 3 ) ^{2}} : \frac{18}{x - 6} - \frac{14}{x - 3} - \frac{14}{( x - 3 ) ^{2}}

= \int \frac{18}{x - 6} - \frac{14}{x - 3} - \frac{14}{( x - 3 ) ^{2}} d x

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{18}{x - 6} d x - \int \frac{14}{x - 3} d x - \int \frac{14}{( x - 3 ) ^{2}} d x

\int \frac{18}{x - 6} d x = 18 ln ∣ x - 6 ∣

\int \frac{14}{x - 3} d x = 14 ln ∣ x - 3 ∣

\int \frac{14}{( x - 3 ) ^{2}} d x = - \frac{14}{x - 3}

= 18 ln ∣ x - 6 ∣ - 14 ln ∣ x - 3 ∣ - (- \frac{14}{x - 3})

Simplify

= 18 ln ∣ x - 6 ∣ - 14 ln ∣ x - 3 ∣ + \frac{14}{x - 3}

Add a constant to the solution

= 18 ln ∣ x - 6 ∣ - 14 ln ∣ x - 3 ∣ + \frac{14}{x - 3} + C

f (x) = \frac{2}{x + 1}

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

x \to - 2 lim (\frac{x}{2 x ^{2} + 2 x + 1})

\int 1 2^{x} d x

(ln (1 - x))^{^{'}}

What is the integral of (4x^2+4x-6)/((x-6)(x-3)^2) ?
The integral of (4x^2+4x-6)/((x-6)(x-3)^2) is 18ln|x-6|-14ln|x-3|+(14)/(x-3)+C