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Question Number 128346 by Ahmed1hamouda last updated on 06/Jan/21

Answered by MJS_new last updated on 06/Jan/21

$$\int x^2\sqrt{4x^2+1}dx=$$$$[t=2x+\sqrt{4x^2+1}\to dx=\frac{\sqrt{4x^2+1}}{2(2x+\sqrt{4x^2+1})}dt]$$$$=\int (\frac{t^3}{128}-\frac{1}{64t}+\frac{1}{128t^5})dt=$$$$=\frac{t^4}{512}-\frac{\ln t}{64}-\frac{1}{512t^4}=$$$$=\frac{x(8x^2+1)\sqrt{4x^2+1}}{32}-\frac{1}{64}\ln (2x+\sqrt{4x^2+1})+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{9\sqrt{5}}{32}-\frac{1}{64}\ln (2+\sqrt{5})$$

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

$$\mathrm{I}={\int }_0^1\left({\mathrm{x}}^2\sqrt{{4\mathrm{x}}^2+1}\right)\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}2\mathrm{x}=\mathrm{sh}\left(\mathrm{t}\right)\Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{argsh}\left(2\right)}{\left(\frac{\mathrm{sht}}{2}\right)}^2\mathrm{cht}\times \frac{\mathrm{chtdt}}{2}=\frac{1}{8}{\int }_0^{\ln (2+\sqrt{1+2^2})}{\mathrm{sh}}^2\mathrm{t}{\mathrm{ch}}^2\mathrm{t}\mathrm{dt}$$$$=\frac{1}{8}{\int }_0^{\ln (2+\sqrt{5})}{\left(\frac{\mathrm{sh}\left(2\mathrm{t}\right)}{2}\right)}^2\mathrm{dt}=\frac{1}{32}{\int }_0^{\ln (2+\sqrt{5})}\frac{\mathrm{ch}\left(4\mathrm{t}\right)-1}{2}\mathrm{dt}$$$$=\frac{1}{64}{\int }_0^{\ln (2+\sqrt{5})}(\mathrm{ch}\left(4\mathrm{t}\right)-1)\mathrm{dt}=\frac{1}{4.64}{\left[\mathrm{ch}\left(4\mathrm{t}\right)\right]}_0^{\ln (2+\sqrt{5})}-\frac{\ln (2+\sqrt{5})}{64}$$$$=\frac{1}{4.64}{\left[\frac{{\mathrm{e}}^{4\mathrm{t}}+{\mathrm{e}}^{-4\mathrm{t}}}{2}\right]}_0^{\ln (2+\sqrt{5})}-\frac{\ln (2+\sqrt{5})}{64}$$$$=\frac{1}{8.64}\{{(2+\sqrt{5})}^4+{(2+\sqrt{5})}^{-4}-2\}$$

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