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

The integral of sqrt(3-2x)x^2 is -1/35 (5x^2+6x+6)(-2x+3)^{3/2}+C

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

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

- \frac{1}{35} (5 x^{2} + 6 x + 6) (- 2 x + 3)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{u ( u - 3 ) ^{2}}{8} d u

= \int - \frac{u ( u - 3 ) ^{2}}{8} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int u (u - 3)^{2} d u

Expand

u (u - 3)^{2} : u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u

= - \frac{1}{8} \cdot \int u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 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}{8} (\int u^{\frac{5}{2}} d u - \int 6 u^{\frac{3}{2}} d u + \int 9 u d u)

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

\int 6 u^{\frac{3}{2}} d u = \frac{12}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= - \frac{1}{8} (\frac{2}{7} u^{\frac{7}{2}} - \frac{12}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}})

Substitute back

u = 3 - 2 x

= - \frac{1}{8} (\frac{2}{7} (3 - 2 x)^{\frac{7}{2}} - \frac{12}{5} (3 - 2 x)^{\frac{5}{2}} + 6 (3 - 2 x)^{\frac{3}{2}})

Simplify

- \frac{1}{8} (\frac{2}{7} (3 - 2 x)^{\frac{7}{2}} - \frac{12}{5} (3 - 2 x)^{\frac{5}{2}} + 6 (3 - 2 x)^{\frac{3}{2}}) : - \frac{1}{35} (5 x^{2} + 6 x + 6) (- 2 x + 3)^{\frac{3}{2}}

= - \frac{1}{35} (5 x^{2} + 6 x + 6) (- 2 x + 3)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{35} (5 x^{2} + 6 x + 6) (- 2 x + 3)^{\frac{3}{2}} + C

x = 7 y^{2}, x = 5 + 2 y^{2}

f (x) = 4 x, at x = 25

\frac{\partial}{\partial x} ((t + 1)^{3})

\frac{d}{d x} (sin (cos (x^{2})))

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

What is the integral of sqrt(3-2x)x^2 ?
The integral of sqrt(3-2x)x^2 is -1/35 (5x^2+6x+6)(-2x+3)^{3/2}+C