Question Number 69167 by rajesh4661kumar@gmail.com last updated on 21/Sep/19
![](https://www.tinkutara.com/question/9422.png)
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](https://www.tinkutara.com/question/Q69197.png)
$${from}\:{tan}^{−\mathrm{1}} \left(\frac{{x}+{y}}{\mathrm{1}−{xy}}\right)={tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} {y} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}+{x}−{x}^{\mathrm{2}} }\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+{x}−\mathrm{1}}{\mathrm{1}−{x}\left({x}−\mathrm{1}\right)}\right)={tan}^{−\mathrm{1}} {x}+{tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {xdx}+\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right){dx} \\ $$$${use}\:{the}\:{comcept}\:{of}\:{integration}\:{by}\:{parts} \\ $$