Question Number 38124 by maxmathsup by imad last updated on 22/Jun/18
$${prove}\:{that}\:\:\int\:\:\:\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+{c} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int\:\:\:\frac{{dx}}{\sqrt{{a}+{x}^{\mathrm{2}} }}\:{with}\:{a}>\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
$${y}={ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \:\right)}\:+{c}\right. \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:}×\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:}\right) \\ $$$$=\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}×\frac{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${dy}=\frac{{dx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${y}=\int\frac{{dx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$