What is the integral of (1-3x)^2 ?

The integral of (1-3x)^2 is -1/9 (1-3x)^3+C

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integral of (1-3x)^2

\int (1 - 3 x)^{2} d x

- \frac{1}{9} (1 - 3 x)^{3} + C

Solution steps

\int (1 - 3 x)^{2} d x

Apply u-substitution

: \int - \frac{u ^{2}}{3} d u

= \int - \frac{u ^{2}}{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^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = 1 - 3 x

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

Simplify

- \frac{1}{3} \cdot \frac{( 1 - 3 x ) ^{3}}{3} : - \frac{1}{9} (1 - 3 x)^{3}

= - \frac{1}{9} (1 - 3 x)^{3}

Add a constant to the solution

= - \frac{1}{9} (1 - 3 x)^{3} + C

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