What is the integral of ((x^2-2)(x^2+6))/(3x^3) ?

The integral of ((x^2-2)(x^2+6))/(3x^3) is 1/3 ((x^2)/2+4ln|x|+6/(x^2))+C

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integral of ((x^2-2)(x^2+6))/(3x^3)

\int \frac{( x ^{2} - 2 ) ( x ^{2} + 6 )}{3 x ^{3}} d x

\frac{1}{3} (\frac{x ^{2}}{2} + 4 ln ∣ x ∣ + \frac{6}{x ^{2}}) + C

Solution steps

\int \frac{( x ^{2} - 2 ) ( x ^{2} + 6 )}{3 x ^{3}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{( x ^{2} - 2 ) ( x ^{2} + 6 )}{x ^{3}} d x

Expand

\frac{( x ^{2} - 2 ) ( x ^{2} + 6 )}{x ^{3}} : x + \frac{4}{x} - \frac{12}{x ^{3}}

= \frac{1}{3} \cdot \int x + \frac{4}{x} - \frac{12}{x ^{3}} 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}{3} (\int x d x + \int \frac{4}{x} d x - \int \frac{12}{x ^{3}} d x)

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int \frac{12}{x ^{3}} d x = - \frac{6}{x ^{2}}

= \frac{1}{3} (\frac{x ^{2}}{2} + 4 ln ∣ x ∣ - (- \frac{6}{x ^{2}}))

Simplify

= \frac{1}{3} (\frac{x ^{2}}{2} + 4 ln ∣ x ∣ + \frac{6}{x ^{2}})

Add a constant to the solution

= \frac{1}{3} (\frac{x ^{2}}{2} + 4 ln ∣ x ∣ + \frac{6}{x ^{2}}) + C

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What is the integral of ((x^2-2)(x^2+6))/(3x^3) ?
The integral of ((x^2-2)(x^2+6))/(3x^3) is 1/3 ((x^2)/2+4ln|x|+6/(x^2))+C