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Question Number 212339 by Spillover last updated on 10/Oct/24

Answered by Ghisom last updated on 10/Oct/24

$$\underset{0}{\stackrel{3}{\int }}\frac{x^{3/4}{(3-x)}^{1/4}}{{(5-x)}^3}dx=$$$$[t={\left(\frac{2x}{5(3-x)}\right)}^{1/4};C=\frac{9\sqrt[4]{2}}{5\sqrt[4]{5}}]$$$$=C\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{t^6}{{(t^4+1)}^3}dt=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{C}{32}{\left[\frac{t^3(3t^4-1)}{{(t^4+1)}^2}\right]}_0^{\mathrm{\infty }}+\frac{3C}{32}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{t^2}{t^4+1}dt=$$$$\left[\mathrm{use}\mathrm{common}\mathrm{method}\right]$$$$=\frac{27\pi }{320\sqrt[4]{10}}$$

Answered by Ghisom last updated on 10/Oct/24

$$...\mathrm{or}\mathrm{use}$$$$t=\arcsin \frac{\sqrt{6x}}{3\sqrt{5-x}}$$$$\Rightarrow$$$$\frac{9}{5\sqrt[4]{40}}\underset{0}{\stackrel{\pi /2}{\int }}{\sin }^{5/2}t{\cos }^{3/2}tdt=$$$$=\frac{9}{10\sqrt[4]{40}}\mathrm{B}(\frac{5}{4},\frac{7}{4})=\frac{9}{10\sqrt[4]{40}}\times \frac{3\pi }{16\sqrt{2}}=$$$$=\frac{27\pi }{320\sqrt[4]{10}}$$$$$$

Answered by MathematicalUser2357 last updated on 12/Oct/24

$$0.149060872$$

Commented by Ghisom last updated on 13/Oct/24

$$\mathrm{very}\mathrm{accurate}\mathrm{you}\mathrm{are}$$

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