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Question-69167




Question Number 69167 by rajesh4661kumar@gmail.com last updated on 21/Sep/19
Answered by petrochengula last updated on 21/Sep/19
from tan^(−1) (((x+y)/(1−xy)))=tan^(−1) x+tan^(−1) y  tan^(−1) (((2x−1)/(1+x−x^2 )))=tan^(−1) (((x+x−1)/(1−x(x−1))))=tan^(−1) x+tan^(−1) (x−1)  I=∫_0 ^1 tan^(−1) xdx+∫_0 ^1 tan^(−1) (x−1)dx  use the comcept of integration by parts
$${from}\:{tan}^{−\mathrm{1}} \left(\frac{{x}+{y}}{\mathrm{1}−{xy}}\right)={tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} {y} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}+{x}−{x}^{\mathrm{2}} }\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+{x}−\mathrm{1}}{\mathrm{1}−{x}\left({x}−\mathrm{1}\right)}\right)={tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {xdx}+\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right){dx} \\ $$$${use}\:{the}\:{comcept}\:{of}\:{integration}\:{by}\:{parts} \\ $$

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