What is the integral of (2x^2+8x-8)/(x-4) ?

The integral of (2x^2+8x-8)/(x-4) is x^2+16x+56ln|x-4|-80+C

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integral of (2x^2+8x-8)/(x-4)

\int \frac{2 x ^{2} + 8 x - 8}{x - 4} d x

x^{2} + 16 x + 56 ln ∣ x - 4 ∣ - 80 + C

Solution steps

\int \frac{2 x ^{2} + 8 x - 8}{x - 4} d x

Expand

\frac{2 x ^{2} + 8 x - 8}{x - 4} : \frac{2 x ^{2}}{x - 4} + \frac{8 x}{x - 4} - \frac{8}{x - 4}

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{2 x ^{2}}{x - 4} d x + \int \frac{8 x}{x - 4} d x - \int \frac{8}{x - 4} d x

\int \frac{2 x ^{2}}{x - 4} d x = x^{2} + 8 x + 32 ln ∣ x - 4 ∣ - 48

\int \frac{8 x}{x - 4} d x = 8 (x - 4 + 4 ln ∣ x - 4 ∣)

\int \frac{8}{x - 4} d x = 8 ln ∣ x - 4 ∣

= x^{2} + 8 x + 32 ln ∣ x - 4 ∣ - 48 + 8 (x - 4 + 4 ln ∣ x - 4 ∣) - 8 ln ∣ x - 4 ∣

Simplify

x^{2} + 8 x + 32 ln ∣ x - 4 ∣ - 48 + 8 (x - 4 + 4 ln ∣ x - 4 ∣) - 8 ln ∣ x - 4 ∣ : x^{2} + 16 x + 56 ln ∣ x - 4 ∣ - 80

= x^{2} + 16 x + 56 ln ∣ x - 4 ∣ - 80

Add a constant to the solution

= x^{2} + 16 x + 56 ln ∣ x - 4 ∣ - 80 + C

f (x) = \frac{( x - 2 ) ^{5}}{16}

x \to \infty lim (\frac{2 x ^{3} - 5 x + 4}{x ^{2} + 3 x - 4})

x cos (x)

\int \frac{9}{x ^{2} - 6 x + 18} d x

\frac{0.6}{s}

What is the integral of (2x^2+8x-8)/(x-4) ?
The integral of (2x^2+8x-8)/(x-4) is x^2+16x+56ln|x-4|-80+C