What is the integral of 3xsqrt(3-3x) ?

The integral of 3xsqrt(3-3x) is (6sqrt(3))/5 (1-x)^{5/2}-2sqrt(3)(1-x)^{3/2}+C

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integral of 3xsqrt(3-3x)

\int 3 x 3 - 3 x d x

\frac{6 3}{5} (1 - x)^{\frac{5}{2}} - 23 (1 - x)^{\frac{3}{2}} + C

Solution steps

\int 3 x 3 - 3 x d x

Take the constant out:

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

= 3 \cdot \int x 3 - 3 x d x

x 3 - 3 x = 3 x 1 - x

= 3 \cdot \int 3 x 1 - x d x

Take the constant out:

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

= 33 \cdot \int x 1 - x d x

Apply u-substitution

: \int - (- u + 1) u d u

= 33 \cdot \int - (- u + 1) u d u

Take the constant out:

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

= 33 (- \int (- u + 1) u d u)

Expand

(- u + 1) u : - u^{\frac{3}{2}} + u

= 33 (- \int - u^{\frac{3}{2}} + u d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 33 (- (- \int u^{\frac{3}{2}} d u + \int u d u))

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= 33 (- (- \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}}))

Substitute back

u = 1 - x

= 33 (- (- \frac{2}{5} (1 - x)^{\frac{5}{2}} + \frac{2}{3} (1 - x)^{\frac{3}{2}}))

Simplify

33 (- (- \frac{2}{5} (1 - x)^{\frac{5}{2}} + \frac{2}{3} (1 - x)^{\frac{3}{2}})) : \frac{6 3}{5} (1 - x)^{\frac{5}{2}} - 23 (1 - x)^{\frac{3}{2}}

= \frac{6 3}{5} (1 - x)^{\frac{5}{2}} - 23 (1 - x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{6 3}{5} (1 - x)^{\frac{5}{2}} - 23 (1 - x)^{\frac{3}{2}} + C

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What is the integral of 3xsqrt(3-3x) ?
The integral of 3xsqrt(3-3x) is (6sqrt(3))/5 (1-x)^{5/2}-2sqrt(3)(1-x)^{3/2}+C