Question Number 69167 by rajesh4661kumar@gmail.com last updated on 21/Sep/19
Answered by petrochengula last updated on 21/Sep/19
$${from}\:{tan}^{−\mathrm{1}} \left(\frac{{x}+{y}}{\mathrm{1}−{xy}}\right)={tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} {y} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}+{x}−{x}^{\mathrm{2}} }\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+{x}−\mathrm{1}}{\mathrm{1}−{x}\left({x}−\mathrm{1}\right)}\right)={tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {xdx}+\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right){dx} \\ $$$${use}\:{the}\:{comcept}\:{of}\:{integration}\:{by}\:{parts} \\ $$