Question Number 152887 by otchereabdullai@gmail.com last updated on 02/Sep/21
$$\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\:\frac{\mathrm{3}}{\:\sqrt{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{3}} }} \\ $$
Answered by Ar Brandon last updated on 02/Sep/21
![I=∫_1 ^2 (3/( (√((x^2 +3)^3 ))))dx =3∫_1 ^2 (dx/(x^3 (√((1+(3/x^2 ))^3 )))), u=1+(3/x^2 )⇒du=−(6/x^3 )dx =−(1/2)∫_4 ^(7/4) (du/( (√u^3 )))=−(1/2)∫_4 ^(7/4) u^(−(3/2)) du=[u^(−(1/2)) ]_4 ^(7/4) =((√(4/7))−2)=2((1/( (√7)))−1)](Q152893.png)
$${I}=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{3}}{\:\sqrt{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{3}} }}{dx} \\ $$$$\:\:=\mathrm{3}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dx}}{{x}^{\mathrm{3}} \sqrt{\left(\mathrm{1}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)^{\mathrm{3}} }},\:{u}=\mathrm{1}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\Rightarrow{du}=−\frac{\mathrm{6}}{{x}^{\mathrm{3}} }{dx} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{4}} ^{\frac{\mathrm{7}}{\mathrm{4}}} \frac{{du}}{\:\sqrt{{u}^{\mathrm{3}} }}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{4}} ^{\frac{\mathrm{7}}{\mathrm{4}}} {u}^{−\frac{\mathrm{3}}{\mathrm{2}}} {du}=\left[{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right]_{\mathrm{4}} ^{\frac{\mathrm{7}}{\mathrm{4}}} \\ $$$$\:\:=\left(\sqrt{\frac{\mathrm{4}}{\mathrm{7}}}−\mathrm{2}\right)=\mathrm{2}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}−\mathrm{1}\right) \\ $$