Question Number 134182 by Algoritm last updated on 28/Feb/21
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Answered by Ñï= last updated on 01/Mar/21
![∫_0 ^1 (tan^(−1) ((1−x)/(1+x)))^3 (((1+x))/((1−x)(1+x^2 )))dx =∫_0 ^1 (tan^(−1) ((1−x)/(1+x)))^3 ((1+x)/(1−x))dtan^(−1) x =∫_0 ^1 (tan^(−1) ((1−x)/(1+x)))^3 ((1+x)/(1−x))d(tan^(−1) 1−tan^(−1) ((1−x)/(1+x))) =−∫_0 ^1 (tan^(−1) ((1−x)/(1+x)))^3 ((1+x)/(1−x))d(tan^(−1) ((1−x)/(1+x))) =−(1/4)∫_0 ^1 ((1+x)/(1−x))d(tan^(−1) ((1−x)/(1+x)))^4 ....continue.....](https://www.tinkutara.com/question/Q134206.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{3}} \frac{\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{3}} \frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}{d}\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{3}} \frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}{d}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{1}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right) \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{3}} \frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}{d}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}{d}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{4}} \\ $$$$….{continue}….. \\ $$