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Question Number 21394 by mondodotto@gmail.com last updated on 22/Sep/17

Answered by mrW1 last updated on 22/Sep/17

$$\mathrm{t}=\sqrt{2\mathrm{x}+1}$$$$\mathrm{x}=\frac{{\mathrm{t}}^2-1}{2}$$$$\mathrm{dx}=\mathrm{tdt}$$$$\int \mathrm{x}\sqrt{2\mathrm{x}+1}\mathrm{dx}=\int \frac{{\mathrm{t}}^2-1}{2}\times \mathrm{t}\times \mathrm{tdt}$$$$=\frac{1}{2}\int ({\mathrm{t}}^4-{\mathrm{t}}^2)\mathrm{dt}$$$$=\frac{{\mathrm{t}}^5}{10}-\frac{{\mathrm{t}}^3}{6}+\mathrm{C}$$$$=\frac{{\mathrm{t}}^3}{2}(\frac{{\mathrm{t}}^2}{5}-\frac{1}{3})+\mathrm{C}$$$$=\frac{(2\mathrm{x}+1)\sqrt{2\mathrm{x}+1}}{2}[\frac{2\mathrm{x}+1}{5}-\frac{1}{3}]+\mathrm{C}$$$$=\frac{(3\mathrm{x}-1)(2\mathrm{x}+1)\sqrt{2\mathrm{x}+1}}{15}+\mathrm{C}$$

Answered by sma3l2996 last updated on 22/Sep/17

$$t=\sqrt{2x+1}\Rightarrow tdt=dx$$$$\int x\sqrt{2x+1}dx=\int \left(\frac{t^2-1}{2}\right)\times t\left(tdt\right)=\frac{1}{2}\int (t^4-t^2)dt$$$$=\frac{t^3}{2}(\frac{1}{5}{t}^2-\frac{1}{3})+C$$$$=\frac{1}{2}\sqrt{{(2x+1)}^3}(\frac{2x+1}{5}-\frac{1}{3})+C$$$$\int x\sqrt{2x+1}dx=\frac{1}{30}\sqrt{{(2x+1)}^3}(6x-2)+C$$

Commented by edward last updated on 23/Sep/17

$$Kigunduscheckspapa$$

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