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Category: Integration

tan-1-x-1-x-1-dx-

Question Number 17506 by tawa tawa last updated on 06/Jul/17 tan1(x+1x1)dx Answered by sma3l2996 last updated on 06/Jul/17 $${u}={tan}^{−\mathrm{1}} \left(\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\right)\Rightarrow{u}'=\frac{−\mathrm{1}}{\mathrm{2}{x}\sqrt{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}}=\frac{−\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \