What is the integral of (x^3-4x^2+1)/(x-4) ?

The integral of (x^3-4x^2+1)/(x-4) is (x^3)/3+ln|x-4|+96+C

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integral of (x^3-4x^2+1)/(x-4)

\int \frac{x ^{3} - 4 x ^{2} + 1}{x - 4} d x

\frac{x ^{3}}{3} + ln ∣ x - 4 ∣ + 96 + C

Solution steps

\int \frac{x ^{3} - 4 x ^{2} + 1}{x - 4} d x

Expand

\frac{x ^{3} - 4 x ^{2} + 1}{x - 4} : \frac{x ^{3}}{x - 4} - \frac{4 x ^{2}}{x - 4} + \frac{1}{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{4 x ^{2}}{x - 4} d x + \int \frac{1}{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{4 x ^{2}}{x - 4} d x = 4 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣)

\int \frac{1}{x - 4} d x = ln ∣ x - 4 ∣

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

Simplify

\frac{x ^{3}}{3} + 2 x^{2} + 16 x + 64 ln ∣ x - 4 ∣ - 4 (\frac{( x - 4 ) ^{2}}{2} + 8 (x - 4) + 16 ln ∣ x - 4 ∣) + ln ∣ x - 4 ∣ : \frac{x ^{3}}{3} + ln ∣ x - 4 ∣ + 96

= \frac{x ^{3}}{3} + ln ∣ x - 4 ∣ + 96

Add a constant to the solution

= \frac{x ^{3}}{3} + ln ∣ x - 4 ∣ + 96 + C

What is the integral of (x^3-4x^2+1)/(x-4) ?
The integral of (x^3-4x^2+1)/(x-4) is (x^3)/3+ln|x-4|+96+C