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Question Number 56392 by otchereabdullai@gmail.com last updated on 15/Mar/19

$$\mathrm{find}\mathrm{the}\mathrm{integral}\mathrm{of}\int \mathrm{x}\sqrt{\mathrm{x}-1}\mathrm{dx}$$

Commented by maxmathsup by imad last updated on 16/Mar/19

$$changement\sqrt{x-1}=tgivex-1=t^2\Rightarrow$$$$\int x\sqrt{x-1}dx=\int (1+t^2)t\left(2t\right)dt=2\int (t^2+t^4)dt=\frac{2}{3}{t}^3+\frac{2}{5}{t}^5+c$$$$=\frac{2}{3}{\left(\sqrt{x-1}\right)}^3+\frac{2}{5}{\left(\sqrt{x-1)}\right)}^5+c.$$$$=\frac{2}{3}(x-1)\sqrt{x-1}+\frac{2}{5}{(x-1)}^2\sqrt{x-1}+c.$$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}!$$

Answered by kaivan.ahmadi last updated on 15/Mar/19

$$u=x-1\Rightarrow du=dxandx=u+1$$$$\int (u+1)\sqrt{u}du=\int u\sqrt{u}du+\int \sqrt{u}du=$$$$\int u^{\frac{3}{2}}du+\int u^{\frac{1}{2}}du=\frac{2u^{\frac{5}{2}}}{5}+\frac{2u^{\frac{3}{2}}}{3}+C=$$$$\frac{2{(x-1)}^2\sqrt{x-1}}{5}+\frac{2(x-1)\sqrt{x-1}}{3}+C$$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by MJS last updated on 15/Mar/19

$$\int x\sqrt{x-1}dx=$$$$[t=x-1\to dx=dt]$$$$=\int (t+1)\sqrt{t}dt=\int t^{\frac{3}{2}}dt+\int t^{\frac{1}{2}}dt=\frac{2}{5}{t}^{\frac{5}{2}}+\frac{2}{3}{t}^{\frac{3}{2}}=$$$$=\frac{2}{15}{t}^{\frac{3}{2}}(3t+5)=\frac{2}{15}(3x+2)\sqrt{{(x-1)}^3}+C$$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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