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Question Number 128370 by mathocean1 last updated on 06/Jan/21

$$f\left(x\right)=(x-1)\sqrt{x-1}on]1;+\mathrm{\infty }[$$$$F\left(x\right)=?$$

Answered by mathmax by abdo last updated on 06/Jan/21

$$\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{derivative}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}-1}+(\mathrm{x}-1)\times \frac{1}{2\sqrt{\mathrm{x}-1}}$$$$=\frac{2(\mathrm{x}-1)+\mathrm{x}-1}{2\sqrt{\mathrm{x}-1}}=\frac{2\mathrm{x}-2+\mathrm{x}-1}{2\sqrt{\mathrm{x}-1}}=\frac{3\mathrm{x}-3}{2\sqrt{\mathrm{x}-1}}$$$$\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{integral}$$$$\mathrm{F}\left(\mathrm{x}\right)=\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int (\mathrm{x}-1)\sqrt{\mathrm{x}-1}\mathrm{dx}{=}_{\sqrt{\mathrm{x}-1}=\mathrm{t}\to \mathrm{x}-1={\mathrm{t}}^2}\int {\mathrm{t}}^2.\mathrm{t}.\left(2\mathrm{t}\right)\mathrm{dt}$$$$=2\int {\mathrm{t}}^4\mathrm{dt}=\frac{2}{5}{\mathrm{t}}^5+\mathrm{C}=\frac{2}{5}{\left(\sqrt{\mathrm{x}-1}\right)}^5+\mathrm{C}$$$$=\frac{2}{5}{(\mathrm{x}-1)}^2\sqrt{\mathrm{x}-1}+\mathrm{C}$$

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