What is the integral of (x^3-4x^2+5)/(x-4) ?

The integral of (x^3-4x^2+5)/(x-4) is (x^3)/3+5ln|x-4|+C

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integral of (x^3-4x^2+5)/(x-4)

\int \frac{x ^{3} - 4 x ^{2} + 5}{x - 4} d x

\frac{x ^{3}}{3} + 5 ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x ^{3} - 4 x ^{2} + 5}{x - 4} d x

Long division

\frac{x ^{3} - 4 x ^{2} + 5}{x - 4} : x^{2} + \frac{5}{x - 4}

= \int x^{2} + \frac{5}{x - 4} 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 x^{2} d x + \int \frac{5}{x - 4} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int \frac{5}{x - 4} d x = 5 ln ∣ x - 4 ∣

= \frac{x ^{3}}{3} + 5 ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + 5 ln ∣ x - 4 ∣ + C

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What is the integral of (x^3-4x^2+5)/(x-4) ?
The integral of (x^3-4x^2+5)/(x-4) is (x^3)/3+5ln|x-4|+C