Differentiation Questions

Question Number 167554 by cortano1 last updated on 19/Mar/22

$$\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3x}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{5x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }}\:\mathrm{dx} \\$$ $$\\$$

Answered by MJS_new last updated on 19/Mar/22

$$\int\frac{\mathrm{3}{x}^{\mathrm{2}} }{\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{dx}= \\$$ $$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{5}}{x}+\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{\mathrm{5}}{t}}\right] \\$$ $$=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{25}}\int\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\$$ $$=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{25}}\int\frac{{dt}}{{t}}−\frac{\mathrm{12}\sqrt{\mathrm{5}}}{\mathrm{25}}\int\frac{{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\$$ $$=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{25}}\mathrm{ln}\:{t}\:+\frac{\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{25}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}= \\$$ $$=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{25}}\mathrm{ln}\:\left(\sqrt{\mathrm{5}}{x}+\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{1}}\right)\:−\frac{\mathrm{3}{x}}{\mathrm{5}\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{1}}}+{C} \\$$

Commented bycortano1 last updated on 19/Mar/22

$$\mathrm{I}\:\mathrm{got}\:\mathrm{diverges}.\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}? \\$$

Commented byMJS_new last updated on 19/Mar/22

$$\mathrm{yes} \\$$

Commented bycortano1 last updated on 19/Mar/22

$$\mathrm{thanks}\:\mathrm{sir} \\$$

Answered by greogoury55 last updated on 19/Mar/22

$$\:{set}\:{x}\sqrt{\mathrm{5}}\:=\:\mathrm{tan}\:{y}\: \\$$ $$\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{3}\left(\frac{\mathrm{tan}\:^{\mathrm{2}} {y}}{\mathrm{5}}\right)}{\:\left(\sqrt{\mathrm{sec}\:^{\mathrm{2}} {y}}\right)^{\mathrm{3}} }\left(\frac{\mathrm{sec}\:^{\mathrm{2}} {y}}{\:\sqrt{\mathrm{5}}}\right){dy}\: \\$$ $$\:{I}=\:\frac{\mathrm{3}}{\mathrm{5}\sqrt{\mathrm{5}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{tan}\:^{\mathrm{2}} {y}}{\mathrm{sec}\:{y}}\:{dy}\: \\$$ $$\:{I}=\:\frac{\mathrm{3}}{\mathrm{5}\sqrt{\mathrm{5}}}\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {y}}{\mathrm{cos}\:{y}}\:{dy} \\$$ $$\:{I}=\:\frac{\mathrm{3}}{\mathrm{5}\sqrt{\mathrm{5}}}\:\left[\:\mathrm{ln}\:\mid\mathrm{sec}\:{y}+\mathrm{tan}\:{y}\mid−\mathrm{sin}\:{y}\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$ $$\:{Diverges} \\$$

Answered by Mathspace last updated on 20/Mar/22

$$\sqrt{\mathrm{5}}{x}={tant}\:\Rightarrow{I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{3}\left(\frac{{tant}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{tan}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\$$ $$=\frac{\mathrm{3}}{\mathrm{5}\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{tan}^{\mathrm{2}} {t}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} }}{dt} \\$$ $$=\frac{\mathrm{3}}{\mathrm{5}\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{tan}^{\mathrm{2}} {t}−\mathrm{1}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}{dt} \\$$ $$=\frac{\mathrm{3}}{\mathrm{5}\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}{dt}−\frac{\mathrm{3}}{\mathrm{5}\sqrt{\mathrm{5}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dt}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}} \\$$ $${we}\:{have}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}{dt} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dt}}{{cost}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={y}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{dy}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{dy}}{\mathrm{1}−{y}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}−{y}}+\frac{\mathrm{1}}{\mathrm{1}+{y}}\right){dy} \\$$ $$=\left[{ln}\mid\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} =\infty \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} }}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cost}\:{dt} \\$$ $$=\left[{sint}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{1}\:{so}\:{this}\:{integrale}\:{is} \\$$ $${divdrgente}...! \\$$