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

The integral of sqrt((3x-4)^3) is 2/15 (3x-4)^{5/2}+C

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integral of sqrt((3x-4)^3)

integral

(3 x - 4)^{3}

\frac{2}{15} (3 x - 4)^{\frac{5}{2}} + C

Solution steps

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

Apply u-substitution

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

= \int \frac{u ^{\frac{3}{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^{\frac{3}{2}} d u

Apply the Power Rule

: \frac{2}{5} u^{\frac{5}{2}}

= \frac{1}{3} \cdot \frac{2}{5} u^{\frac{5}{2}}

Substitute back

u = 3 x - 4

= \frac{1}{3} \cdot \frac{2}{5} (3 x - 4)^{\frac{5}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{5} (3 x - 4)^{\frac{5}{2}} : \frac{2}{15} (3 x - 4)^{\frac{5}{2}}

= \frac{2}{15} (3 x - 4)^{\frac{5}{2}}

Add a constant to the solution

= \frac{2}{15} (3 x - 4)^{\frac{5}{2}} + C

sin^{2} (3 x)

x^{2} + a^{2}

\frac{y ^{2}}{2}

arctan (\frac{1 - 1}{x})

\frac{x}{x ^{2} + a ^{2}}

What is the integral of sqrt((3x-4)^3) ?
The integral of sqrt((3x-4)^3) is 2/15 (3x-4)^{5/2}+C