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

The integral of x^2sqrt(2-3x) is -1/27 (2/7 (2-3x)^{7/2}-8/5 (2-3x)^{5/2}+8/3 (2-3x)^{3/2})+C

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integral of x^2sqrt(2-3x)

\int x^{2} 2 - 3 x d x

- \frac{1}{27} (\frac{2}{7} (2 - 3 x)^{\frac{7}{2}} - \frac{8}{5} (2 - 3 x)^{\frac{5}{2}} + \frac{8}{3} (2 - 3 x)^{\frac{3}{2}}) + C

Solution steps

\int x^{2} 2 - 3 x d x

Apply u-substitution

: \int - \frac{u ( u - 2 ) ^{2}}{27} d u

= \int - \frac{u ( u - 2 ) ^{2}}{27} d u

Take the constant out:

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

= - \frac{1}{27} \cdot \int u (u - 2)^{2} d u

Expand

u (u - 2)^{2} : u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 u

= - \frac{1}{27} \cdot \int u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 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

= - \frac{1}{27} (\int u^{\frac{5}{2}} d u - \int 4 u^{\frac{3}{2}} d u + \int 4 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 4 u^{\frac{3}{2}} d u = \frac{8}{5} u^{\frac{5}{2}}

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= - \frac{1}{27} (\frac{2}{7} u^{\frac{7}{2}} - \frac{8}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = 2 - 3 x

= - \frac{1}{27} (\frac{2}{7} (2 - 3 x)^{\frac{7}{2}} - \frac{8}{5} (2 - 3 x)^{\frac{5}{2}} + \frac{8}{3} (2 - 3 x)^{\frac{3}{2}})

Add a constant to the solution

= - \frac{1}{27} (\frac{2}{7} (2 - 3 x)^{\frac{7}{2}} - \frac{8}{5} (2 - 3 x)^{\frac{5}{2}} + \frac{8}{3} (2 - 3 x)^{\frac{3}{2}}) + C

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What is the integral of x^2sqrt(2-3x) ?
The integral of x^2sqrt(2-3x) is -1/27 (2/7 (2-3x)^{7/2}-8/5 (2-3x)^{5/2}+8/3 (2-3x)^{3/2})+C