Find $\displaystyle \int{\dfrac{x}{\sqrt{4-x}}\,\mathrm{d}x}$, using the substitution \(u=\sqrt{4-

Question

Find $\displaystyle \int{\dfrac{x}{\sqrt{4-x}}\,\mathrm{d}x}$, using the substitution $u=\sqrt{4-x}$.

Accepted Answer

$-8\sqrt{4-x}+\dfrac{2}{3}(4-x)^{3/2}+C$

Supplied Substitutions