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

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

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integral of 2(1-3x)^2

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

- \frac{2}{9} (1 - 3 x)^{3} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

= 2 (- \frac{1}{3} \cdot \int u^{2} d u)

Apply the Power Rule

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

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

Substitute back

u = 1 - 3 x

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

Simplify

2 (- \frac{1}{3} \cdot \frac{( 1 - 3 x ) ^{3}}{3}) : - \frac{2}{9} (1 - 3 x)^{3}

= - \frac{2}{9} (1 - 3 x)^{3}

Add a constant to the solution

= - \frac{2}{9} (1 - 3 x)^{3} + C

\frac{d}{d x} ((x^{3} - 2)^{2})

\int \frac{1}{e ^{- x} + e ^{x}} d x

t \to 0 + lim (\frac{4}{t} - \frac{1}{e ^{t} - 1})

\frac{d}{d x} (2 csc (x) cot (x))

y = 2 x + 1, at x = \frac{1}{4}

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