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

The integral of (8x^3)/(x-4) is 8((x^3)/3+2x^2+16x+64ln|x-4|)+C

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Long division

\frac{x ^{3}}{x - 4} : x^{2} + 4 x + 16 + \frac{64}{x - 4}

= 8 \cdot \int x^{2} + 4 x + 16 + \frac{64}{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

= 8 (\int x^{2} d x + \int 4 x d x + \int 16 d x + \int \frac{64}{x - 4} d x)

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

\int 4 x d x = 2 x^{2}

\int 16 d x = 16 x

\int \frac{64}{x - 4} d x = 64 ln ∣ x - 4 ∣

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

Add a constant to the solution

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

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What is the integral of (8x^3)/(x-4) ?
The integral of (8x^3)/(x-4) is 8((x^3)/3+2x^2+16x+64ln|x-4|)+C