What is the integral of (4-3x)^2 ?

The integral of (4-3x)^2 is -1/9 (4-3x)^3+C

Popular >

integral of (4-3x)^2

\int (4 - 3 x)^{2} d x

- \frac{1}{9} (4 - 3 x)^{3} + C

Solution steps

\int (4 - 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 = 4 - 3 x

= - \frac{1}{3} \cdot \frac{( 4 - 3 x ) ^{3}}{3}

Simplify

- \frac{1}{3} \cdot \frac{( 4 - 3 x ) ^{3}}{3} : - \frac{1}{9} (4 - 3 x)^{3}

= - \frac{1}{9} (4 - 3 x)^{3}

Add a constant to the solution

= - \frac{1}{9} (4 - 3 x)^{3} + C

\int 2 x^{2} + 7 + \frac{9}{x ^{2} + 1} d x

\frac{d}{d x} ((ln (x))^{3})

\frac{\partial}{\partial x} (\frac{1}{x - b})

f (x) = \frac{( 10 - 3 x ) + 2}{2}

\int \frac{x ^{2}}{x ^{3} + 3} d x