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Question Number 141944 by cesarL last updated on 25/May/21

$$\int x^2\sqrt{9x^2+25}dx$$

Answered by MJS_new last updated on 25/May/21

$$\int x^2\sqrt{9x^2+25}dx=$$$$[t=\frac{\sqrt{9x^2+25}}{x}\to dx=--\frac{x^2\sqrt{9x^2+25}}{25}dt]$$$$=-625\int \frac{t^2}{{(t^2-9)}^3}dt=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{625t(t^2+9)}{72{(t^2-9)}^2}+\frac{625}{72}\int \frac{dt}{t^2-9}=$$$$=\frac{625t(t^2+9)}{72{(t^2-9)}^2}+\frac{625}{432}\int (\frac{1}{t-3}-\frac{1}{t+3})dt=$$$$=\frac{625t(t^2+9)}{72{(t^2-9)}^2}+\frac{625}{432}\ln \frac{t-3}{t+3}=$$$$=\frac{x(18x^2+25)\sqrt{9x^2+25}}{72}-\frac{625}{216}\ln (3x+\sqrt{9x^2+25})+C$$

Answered by MJS_new last updated on 25/May/21

$$\int x^2\sqrt{9x^2+25}dx=$$$$[u=\frac{3x+\sqrt{9x^2+25}}{5}\to dx=\frac{5\sqrt{9x^2+25}}{3(3x+\sqrt{9x^2+25})}du]$$$$=\frac{625}{432}\int \frac{u^8-2u^4+1}{u^5}du=\frac{625}{432}\int (u^3+\frac{1}{u^5}-\frac{2}{u})du=$$$$=\frac{625u^4}{1728}-\frac{625}{1728u^4}-\frac{625}{216}\ln u=$$$$=\frac{x(18x^2+25)\sqrt{9x^2+25}}{72}-\frac{625}{216}\ln (3x+\sqrt{9x^2+25})+C$$

Commented by cesarL last updated on 25/May/21

$$Ineedfortrigonometricsustitution$$$$pleasesir$$

Answered by Ar Brandon last updated on 25/May/21

$$\mathcal{I}=\int {\mathrm{x}}^2\sqrt{{9\mathrm{x}}^2+25}\mathrm{dx}=3\int {\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+{\left(\frac{5}{3}\right)}^2}\mathrm{dx}$$$$\mathrm{x}=\frac{5}{3}\sinh \theta \Rightarrow \mathrm{dx}=\frac{5}{3}\cosh \theta \mathrm{d}\theta$$$$\mathcal{I}=3\int {\left(\frac{5}{3}\sinh \theta \right)}^2\sqrt{{\left(\frac{5}{3}\sinh \theta \right)}^2+{\left(\frac{5}{3}\right)}^2}\cdot \left(\frac{5}{3}\cosh \theta \mathrm{d}\theta \right)$$$$=3\cdot {\left(\frac{5}{3}\right)}^4\int {\sinh }^2\theta \sqrt{{\sinh }^2\theta +1}\cdot \cosh \theta \mathrm{d}\theta$$$$=3{\left(\frac{5}{3}\right)}^4\int {\sinh }^2\theta {\cosh }^2\theta \mathrm{d}\theta =3{\left(\frac{5}{3}\right)}^4\int ({\cosh }^2\theta +{\cosh }^4\theta )\mathrm{d}\theta$$$$=3{\left(\frac{5}{3}\right)}^4\int (\frac{\cosh 2\theta +1}{2}+\frac{\cosh 4\theta +4\cosh 2\theta +3}{8})\mathrm{d}\theta$$

Commented by Ar Brandon last updated on 25/May/21

$$2\cosh \theta ={\mathrm{e}}^{\theta }+{\mathrm{e}}^{-\theta }\Rightarrow {4\cosh }^2\theta ={\mathrm{e}}^{2\theta }+2+{\mathrm{e}}^{-2\theta }=2+2\cosh \theta$$$$\Rightarrow {16\cosh }^4\theta ={\mathrm{e}}^{4\theta }+{4\mathrm{e}}^{2\theta }+6+{4\mathrm{e}}^{-2\theta }+{\mathrm{e}}^{-4\theta }$$$$=2\cosh 4\theta +8\cosh 2\theta +6$$

Answered by mathmax by abdo last updated on 25/May/21

$$\mathrm{\Phi }=\int {\mathrm{x}}^2\sqrt{{9\mathrm{x}}^2+25}\mathrm{dx}\Rightarrow \mathrm{\Phi }=\int {3\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+{\left(\frac{5}{3}\right)}^2}\mathrm{dx}$$$$=3\mathrm{f}\left(\frac{5}{3}\right)\mathrm{with}\mathrm{f}\left(\mathrm{a}\right)=\int {\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{ash}\left(\mathrm{t}\right)\mathrm{give}$$$$\mathrm{f}\left(\mathrm{a}\right)=\int {\mathrm{a}}^2{\mathrm{sh}}^2\left(\mathrm{t}\right)\mathrm{ach}\left(\mathrm{t}\right)\mathrm{acht}\mathrm{dt}={\mathrm{a}}^4\int {\mathrm{sh}}^2\mathrm{t}{\mathrm{ch}}^2\mathrm{t}\mathrm{dt}$$$$={\mathrm{a}}^4\int {\left(\frac{1}{2}\mathrm{sh}\left(2\mathrm{t}\right)\right)}^2\mathrm{dt}=\frac{{\mathrm{a}}^4}{4}\int {\mathrm{sh}}^2\left(2\mathrm{t}\right)\mathrm{dt}$$$$=\frac{{\mathrm{a}}^4}{8}\int (\mathrm{ch}\left(4\mathrm{t}\right)-1)\mathrm{dt}=\frac{{\mathrm{a}}^4}{8}\mathrm{t}-\frac{{\mathrm{a}}^4}{8}\int \mathrm{ch}\left(4\mathrm{t}\right)\mathrm{dt}$$$$=\frac{{\mathrm{a}}^4}{8}\mathrm{t}-\frac{{\mathrm{a}}^4}{64}\mathrm{sh}\left(4\mathrm{t}\right)+\mathrm{C}[\mathrm{we}[\mathrm{have}$$$$\mathrm{sh}\left(4\mathrm{t}\right)=\frac{{\mathrm{e}}^{4\mathrm{t}}+{\mathrm{e}}^{-4\mathrm{t}}}{2}\mathrm{and}\mathrm{t}=\mathrm{argsh}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)=\log (\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})$$$$\Rightarrow \mathrm{sh}\left(4\mathrm{t}\right)=\frac{1}{2}\{{(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^4+{(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^{-4}\}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{{\mathrm{a}}^4}{8}\log (\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})$$$$-\frac{{\mathrm{a}}^4}{128}\{{(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^4+{(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^{-4}\}+\mathrm{C}$$$$\mathrm{\Phi }=3\mathrm{f}\left(\frac{5}{3}\right)$$$$$$

Commented by mathmax by abdo last updated on 25/May/21

$$\mathrm{sorry}\mathrm{f}\left(\mathrm{a}\right)=-\frac{{\mathrm{a}}^4}{8}\mathrm{t}+\frac{{\mathrm{a}}^4}{64}\mathrm{sh}\left(4\mathrm{t}\right)+\mathrm{C}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=-\frac{{\mathrm{a}}^4}{8}\log (\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})+\frac{{\mathrm{a}}^4}{128}\{{(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^4+{(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})}^{-4}\}+\mathrm{C}$$

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