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Question Number 95546 by Fikret last updated on 25/May/20

$$\int x^2\sqrt{a^2+x^2}dx=?$$

Answered by MJS last updated on 25/May/20

$$\int x^2\sqrt{x^2+a^2}dx=$$$$[t=\frac{x+\sqrt{x^2+a^2}}{a}\to dx=\frac{a\sqrt{x^2+a^2}}{x+\sqrt{x^2+a^2}}dt]$$$$=\frac{a^4}{16}\int \frac{t^8-2t^4+1}{t^5}dt=$$$$=\frac{a^4(t^8-1)}{64t^4}-\frac{a^4}{8}\ln t=$$$$=\frac{1}{8}x(2x^2+a^2)\sqrt{x^2+a^2}-\frac{a^4}{8}\ln (x+\sqrt{x^2+a^2})+\mathrm{C}$$

Answered by mathmax by abdo last updated on 26/May/20

$$\mathrm{I}=\int {\mathrm{x}}^2\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{cha}7\mathrm{gement}\mathrm{x}=\mathrm{asht}\Rightarrow$$$$\int {\mathrm{x}}^2\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}\mathrm{dx}=\int {\mathrm{a}}^2{\mathrm{sh}}^2\left(\mathrm{t}\right)\mid \mathrm{a}\mid \mathrm{ch}\left(\mathrm{t}\right)\mathrm{ach}\left(\mathrm{t}\right)\mathrm{dt}$$$$={\mathrm{a}}^3\mid \mathrm{a}\mid \int {\left(\mathrm{sht}\mathrm{cht}\right)}^2\mathrm{dt}=\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{4}\int {\mathrm{sh}}^2\left(2\mathrm{t}\right)\mathrm{dt}=$$$$=\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{4}\int \frac{\mathrm{ch}\left(4\mathrm{t}\right)-1}{2}\mathrm{dt}=\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{8}\int \mathrm{ch}\left(4\mathrm{t}\right)\mathrm{dt}-\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{8}\mathrm{t}$$$$=\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{32}\mathrm{sh}\left(4\mathrm{t}\right)-\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{8}\mathrm{t}+\mathrm{C}$$$$=\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{32}\left(\frac{{\mathrm{e}}^{4\mathrm{t}}-{\mathrm{e}}^{-4\mathrm{t}}}{2}\right)-\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{8}\mathrm{argsh}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$$$$=\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{64}((\ln {(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^4-\left(\ln {(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^{-4}\right)$$$$-\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{8}\ln (\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})+\mathrm{C}$$

Commented by mathmax by abdo last updated on 26/May/20

$$\mathrm{sorry}\mathrm{I}=\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{64}\{{(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^4-{(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^{-4}\}$$$$-\frac{{\mathrm{a}}^3\mid \mathrm{a}\mid }{8}\ln (\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})+\mathrm{C}$$

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