Question Number 38199 by prof Abdo imad last updated on 22/Jun/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\sqrt{{x}}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
$${x}={t}^{\mathrm{2}\:} \:{dx}=\mathrm{2}{tdt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{{ln}\mid\frac{\left({t}+\frac{\mathrm{1}}{{t}}\right)−\sqrt{\mathrm{2}}}{\left({t}+\frac{\mathrm{1}}{{t}}\right)+\sqrt{\mathrm{2}}}\mid+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}}\right)\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$= \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$