Question

o calculo da integral raiz quadrada 3-2x * x² dx. por substituição

Answer 1

Asumiré que la pregunta es
           $\displaystyle \text{Calcular la siguiente integral }\int x^2\sqrt{3-2x}\,dx$

Solución
Sustitución: 
       $u=\sqrt{3-2x}\to du = -\dfrac{x}{\sqrt{3-2x}}dx\to u\,du=-x\,dx$

Por otro lado se ve $u=\sqrt{3-2x}\to x=\dfrac{3-u^2}{2}$, por ello $dx=\dfrac{2u}{u^2-3}du$, entonces sustituimos

        $\displaystyle \int x^2\sqrt{3-2x}\,dx=\int \left(\dfrac{3-u^2}{2}\right)^2\cdot u\cdot \dfrac{2u}{u^2-3}du\\ \\ \\ \int x^2\sqrt{3-2x}\,dx=\dfrac{1}{2}\int u^2(u^2-3)\,du\\ \\ \\ \int x^2\sqrt{3-2x}\,dx=\dfrac{1}{10} u^5-\dfrac{1}{2}u^3+C\\ \\ \\ \boxed{\int x^2\sqrt{3-2x}\,dx=\dfrac{1}{10} \sqrt{3-2x}^{\,5}-\dfrac{1}{2}\sqrt{3-2x}^{\,3}+C}$

Answer 2

Resposta:

The answer above is good but i think theres a little mistake,isn't it?

$u$ = $sqrt(3-2x)$

$du = (\frac{1}{2\sqrt{3-2x} } ) (3-2x)'dx$

$du=(-1/u)dx \\ dx=-udu$

$because \\\frac{d}{dx} (3-2x) = -2$

it will change the integral as follows

$u^{2} = (3-2x)\\ x = \frac{3-u^{2} }{2} \\ \\thus\\\\\int\limits \frac{(3-u^{2})^{2} }{4} u (-u) \, du$