What is the integral of (2-3x)^{2018} ?

The integral of (2-3x)^{2018} is -1/6057 (2-3x)^{2019}+C

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integral of (2-3x)^{2018}

\int (2 - 3 x)^{2018} d x

- \frac{1}{6057} (2 - 3 x)^{2019} + C

Solution steps

\int (2 - 3 x)^{2018} d x

Apply u-substitution

: \int - \frac{u ^{2018}}{3} d u

= \int - \frac{u ^{2018}}{3} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int u^{2018} d u

Apply the Power Rule

: \frac{u ^{2019}}{2019}

= - \frac{1}{3} \cdot \frac{u ^{2019}}{2019}

Substitute back

u = 2 - 3 x

= - \frac{1}{3} \cdot \frac{( 2 - 3 x ) ^{2019}}{2019}

Simplify

- \frac{1}{3} \cdot \frac{( 2 - 3 x ) ^{2019}}{2019} : - \frac{1}{6057} (2 - 3 x)^{2019}

= - \frac{1}{6057} (2 - 3 x)^{2019}

Add a constant to the solution

= - \frac{1}{6057} (2 - 3 x)^{2019} + C

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What is the integral of (2-3x)^{2018} ?
The integral of (2-3x)^{2018} is -1/6057 (2-3x)^{2019}+C