What is the integral of (3/(m^4)-1)(1/(m^5)) ?

The integral of (3/(m^4)-1)(1/(m^5)) is -((3-m^4)^2)/(24m^8)+C

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integral of (3/(m^4)-1)(1/(m^5))

\int (\frac{3}{m ^{4}} - 1) (\frac{1}{m ^{5}}) d m

- \frac{( 3 - m ^{4} ) ^{2}}{24 m ^{8}} + C

Solution steps

\int (\frac{3}{m ^{4}} - 1) \frac{1}{m ^{5}} d m

Apply u-substitution

: \int - \frac{u}{12} d u

= \int - \frac{u}{12} d u

Take the constant out:

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

= - \frac{1}{12} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{1}{12} \cdot \frac{u ^{2}}{2}

Substitute back

u = \frac{3}{m ^{4}} - 1

= - \frac{1}{12} \cdot \frac{( \frac{3}{m ^{4}} - 1 ) ^{2}}{2}

Simplify

- \frac{1}{12} \cdot \frac{( \frac{3}{m ^{4}} - 1 ) ^{2}}{2} : - \frac{( 3 - m ^{4} ) ^{2}}{24 m ^{8}}

= - \frac{( 3 - m ^{4} ) ^{2}}{24 m ^{8}}

Add a constant to the solution

= - \frac{( 3 - m ^{4} ) ^{2}}{24 m ^{8}} + C

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What is the integral of (3/(m^4)-1)(1/(m^5)) ?
The integral of (3/(m^4)-1)(1/(m^5)) is -((3-m^4)^2)/(24m^8)+C