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

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

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integral of (x^3)/(sqrt(x^2)-4)

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

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

Solution steps

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

x^{2} = (x),

assuming

x \geq 0

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

Long division

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

= \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

= \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 ∣

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

Add a constant to the solution

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

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