What is the integral of (4-7m)^{1/5} ?

The integral of (4-7m)^{1/5} is -5/42 (4-7m)^{6/5}+C

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integral of (4-7m)^{1/5}

\int (4 - 7 m)^{\frac{1}{5}} d m

- \frac{5}{42} (4 - 7 m)^{\frac{6}{5}} + C

Solution steps

\int (4 - 7 m)^{\frac{1}{5}} d m

Apply u-substitution

: \int - \frac{u ^{\frac{1}{5}}}{7} d u

= \int - \frac{u ^{\frac{1}{5}}}{7} d u

Take the constant out:

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

= - \frac{1}{7} \cdot \int u^{\frac{1}{5}} d u

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= - \frac{1}{7} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 4 - 7 m

= - \frac{1}{7} \cdot \frac{5}{6} (4 - 7 m)^{\frac{6}{5}}

Simplify

- \frac{1}{7} \cdot \frac{5}{6} (4 - 7 m)^{\frac{6}{5}} : - \frac{5}{42} (4 - 7 m)^{\frac{6}{5}}

= - \frac{5}{42} (4 - 7 m)^{\frac{6}{5}}

Add a constant to the solution

= - \frac{5}{42} (4 - 7 m)^{\frac{6}{5}} + C

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What is the integral of (4-7m)^{1/5} ?
The integral of (4-7m)^{1/5} is -5/42 (4-7m)^{6/5}+C