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

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

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

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

\frac{m ^{2}}{6} + ln ∣ m ∣ + \frac{3}{m ^{2}} + C

Solution steps

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

Take the constant out:

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

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

Expand

\frac{( m ^{2} - 3 ) ( m ^{2} + 6 )}{m ^{3}} : m + \frac{3}{m} - \frac{18}{m ^{3}}

= \frac{1}{3} \cdot \int m + \frac{3}{m} - \frac{18}{m ^{3}} d m

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 m d m + \int \frac{3}{m} d m - \int \frac{18}{m ^{3}} d m)

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

\int \frac{3}{m} d m = 3 ln ∣ m ∣

\int \frac{18}{m ^{3}} d m = - \frac{9}{m ^{2}}

= \frac{1}{3} (\frac{m ^{2}}{2} + 3 ln ∣ m ∣ - (- \frac{9}{m ^{2}}))

Simplify

\frac{1}{3} (\frac{m ^{2}}{2} + 3 ln ∣ m ∣ - (- \frac{9}{m ^{2}})) : \frac{m ^{2}}{6} + ln ∣ m ∣ + \frac{3}{m ^{2}}

= \frac{m ^{2}}{6} + ln ∣ m ∣ + \frac{3}{m ^{2}}

Add a constant to the solution

= \frac{m ^{2}}{6} + ln ∣ m ∣ + \frac{3}{m ^{2}} + C

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