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Question Number 63433 by minh2001 last updated on 04/Jul/19

$$\underset{1}{\stackrel{x}{\int }}{x}^2-3x\sqrt{x}dx=\frac{-716}{15}$$$$thencalculate\underset{x}{\stackrel{x+1}{\int }}\frac{1}{x+3}dx$$

Commented by MJS last updated on 04/Jul/19

$$\mathrm{the}\mathrm{borders}\mathrm{have}\mathrm{to}\mathrm{be}\mathrm{independent}\mathrm{of}\mathrm{the}$$$$\mathrm{integral}\mathrm{variable}$$

Answered by MJS last updated on 04/Jul/19

$$\underset{1}{\stackrel{t}{\int }}(x^2-3x\sqrt{x})dx=-\frac{716}{15}$$$$\frac{1}{3}{t}^3-\frac{6}{5}{t}^{\frac{5}{2}}+\frac{13}{15}=-\frac{716}{15}$$$$t^3-\frac{18}{5}{t}^{\frac{5}{2}}+\frac{729}{5}=0$$$$t=u^2$$$$u^6-\frac{18}{5}{u}^5+\frac{729}{5}=0$$$$\Rightarrow u_1=u_2=3\Rightarrow t=9$$$$\underset{9}{\stackrel{10}{\int }}\frac{dx}{x+3}=\ln \frac{13}{12}$$

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