What is the integral of (x^3-3x-8)/(x-4) ?

The integral of (x^3-3x-8)/(x-4) is (x^3)/3+2x^2+13x+44ln|x-4|+12+C

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integral of (x^3-3x-8)/(x-4)

\int \frac{x ^{3} - 3 x - 8}{x - 4} d x

\frac{x ^{3}}{3} + 2 x^{2} + 13 x + 44 ln ∣ x - 4 ∣ + 12 + C

Solution steps

\int \frac{x ^{3} - 3 x - 8}{x - 4} d x

Expand

\frac{x ^{3} - 3 x - 8}{x - 4} : \frac{x ^{3}}{x - 4} - \frac{3 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{x ^{3}}{x - 4} d x - \int \frac{3 x}{x - 4} d x - \int \frac{8}{x - 4} d x

\int \frac{x ^{3}}{x - 4} d x = \frac{x ^{3}}{3} + 2 x^{2} + 16 x + 64 ln ∣ x - 4 ∣

\int \frac{3 x}{x - 4} d x = 3 (x - 4 + 4 ln ∣ x - 4 ∣)

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

= \frac{x ^{3}}{3} + 2 x^{2} + 16 x + 64 ln ∣ x - 4 ∣ - 3 (x - 4 + 4 ln ∣ x - 4 ∣) - 8 ln ∣ x - 4 ∣

Simplify

\frac{x ^{3}}{3} + 2 x^{2} + 16 x + 64 ln ∣ x - 4 ∣ - 3 (x - 4 + 4 ln ∣ x - 4 ∣) - 8 ln ∣ x - 4 ∣ : \frac{x ^{3}}{3} + 2 x^{2} + 13 x + 44 ln ∣ x - 4 ∣ + 12

= \frac{x ^{3}}{3} + 2 x^{2} + 13 x + 44 ln ∣ x - 4 ∣ + 12

Add a constant to the solution

= \frac{x ^{3}}{3} + 2 x^{2} + 13 x + 44 ln ∣ x - 4 ∣ + 12 + C

\int \frac{1}{x ^{4} + 16 x ^{2}} d x

\int \frac{x ^{2}}{2 x - x ^{2}} d x

\int \frac{x + 3}{6 x ^{3} - 7 x ^{2} - 3 x} d x

\int \frac{1}{a x - x ^{2}} d x

\int \frac{1}{x ^{2} - a x} d x

What is the integral of (x^3-3x-8)/(x-4) ?
The integral of (x^3-3x-8)/(x-4) is (x^3)/3+2x^2+13x+44ln|x-4|+12+C