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

The integral of ((x^2-4)^2)/(2x) is 1/2 ((x^4)/4-4x^2+16ln|x|)+C

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

\int \frac{( x ^{2} - 4 ) ^{2}}{2 x} d x

\frac{1}{2} (\frac{x ^{4}}{4} - 4 x^{2} + 16 ln ∣ x ∣) + C

Solution steps

\int \frac{( x ^{2} - 4 ) ^{2}}{2 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{( x ^{2} - 4 ) ^{2}}{x} d x

Expand

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

= \frac{1}{2} \cdot \int x^{3} - 8 x + \frac{16}{x} 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

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

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

\int 8 x d x = 4 x^{2}

\int \frac{16}{x} d x = 16 ln ∣ x ∣

= \frac{1}{2} (\frac{x ^{4}}{4} - 4 x^{2} + 16 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{4}}{4} - 4 x^{2} + 16 ln ∣ x ∣) + C

What is the integral of ((x^2-4)^2)/(2x) ?
The integral of ((x^2-4)^2)/(2x) is 1/2 ((x^4)/4-4x^2+16ln|x|)+C