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Question Number 105386 by bemath last updated on 28/Jul/20

$$\int \frac{x^3}{\sqrt{{(a^2+x^2)}^3}}dx$$

Answered by john santu last updated on 28/Jul/20

$$bysubstitutex=a\tan \theta$$$$I=\int \frac{x^{3}dx}{\sqrt{{(a^2+x^2)}^3}}=a\int \frac{\tan {}^3\theta }{\sec {}^3\theta }.{\sec }^2\theta d\theta$$$$I=a\int \frac{\tan {}^3\theta }{\sec \theta }d\theta =a\int \frac{\sin {}^3\theta }{\cos {}^2\theta }d\theta$$$$setw=\cos \theta$$$$I=-a\int \frac{1-w^2}{w^2}dw=aw+\frac{a}{w}+c$$$$\therefore I=a\cos \theta +\frac{a}{\cos \theta }+c$$$$I=\frac{a^2}{\sqrt{a^2+x^2}}+\sqrt{a^2+x^2}+c$$$$\left(JS\mathrm{♠}⧫\right)$$

Answered by mathmax by abdo last updated on 28/Jul/20

$$\mathrm{I}=\int \frac{{\mathrm{x}}^3}{\sqrt{{({\mathrm{a}}^2+{\mathrm{x}}^2)}^3}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{asht}\Rightarrow$$$$\mathrm{I}=\int \frac{{\mathrm{a}}^3{\mathrm{sh}}^3\mathrm{t}}{\sqrt{{\mathrm{a}}^6{\left(\mathrm{cht}\right)}^6}}\mathrm{ach}\left(\mathrm{t}\right)\mathrm{dt}=\int \frac{{\mathrm{a}}^3{\mathrm{sh}}^3\mathrm{t}}{{\mathrm{a}}^3{\mathrm{ch}}^3\mathrm{t}}\times \mathrm{ach}\left(\mathrm{t}\right)\mathrm{dt}$$$$=\mathrm{a}\int \frac{{\mathrm{sh}}^3\mathrm{t}}{{\mathrm{ch}}^2\mathrm{t}}\mathrm{dt}=\mathrm{a}\int \frac{{\mathrm{sh}}^2\mathrm{t}\mathrm{sht}}{{\mathrm{ch}}^2\mathrm{t}}\mathrm{dt}=\mathrm{a}\int \frac{\mathrm{sht}({\mathrm{ch}}^2\mathrm{t}-1)}{{\mathrm{ch}}^2\mathrm{t}}\mathrm{dt}$$$$=\mathrm{a}\int \mathrm{sh}\left(\mathrm{t}\right)-\mathrm{a}\int \frac{\mathrm{sht}}{{\mathrm{ch}}^2\mathrm{t}}\mathrm{dt}=\mathrm{ach}\left(\mathrm{t}\right)+\frac{\mathrm{a}}{\mathrm{ch}\left(\mathrm{t}\right)}+\mathrm{c}$$$$\mathrm{but}\mathrm{t}=\mathrm{argsh}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)=\ln (\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}})\mathrm{and}\mathrm{ch}\left(\mathrm{t}\right)=\sqrt{1+{\mathrm{sh}}^2\mathrm{t}}$$$$=\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}}\Rightarrow \mathrm{I}=\mathrm{a}\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}}+\frac{\mathrm{a}}{\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}}}+\mathrm{C}$$

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