Question Number 89596 by A8;15: last updated on 18/Apr/20
Commented by john santu last updated on 18/Apr/20
![(√(x/(1−x))) = t ⇒ x = (t^2 /(t^2 +1)) dx = ((2t)/((t^2 +1)^2 )) dt ] ∫ ((2t^2 )/((t^2 +1)^2 )) dt = ∫ t(((d(t^2 +1))/((t^2 +1)^2 ))) = [ by parts ] −(t/(t^2 +1)) + ∫ (dt/(t^2 +1)) = −(t/(t^2 +1)) + tan^(−1) (t) + c = −((√(x/(1−x)))/(1/(1−x))) + tan^(−1) ((√(x/(1−x)))) + c −(√(x−x^2 )) + tan^(−1) ((√(x/(1−x)))) + c](Q89601.png)
$$\sqrt{\frac{{x}}{\mathrm{1}−{x}}}\:=\:{t}\:\Rightarrow\:{x}\:=\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\left.{dx}\:=\:\frac{\mathrm{2}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dt}\:\right]\: \\ $$$$\int\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dt}\:= \\ $$$$\int\:{t}\left(\frac{{d}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right)\:=\:\left[\:{by}\:{parts}\:\right] \\ $$$$−\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:+\:\int\:\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\:=\: \\ $$$$−\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:+\:\mathrm{tan}^{−\mathrm{1}} \left({t}\right)\:+\:{c}\:=\: \\ $$$$−\frac{\sqrt{\frac{{x}}{\mathrm{1}−{x}}}}{\frac{\mathrm{1}}{\mathrm{1}−{x}}}\:+\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\frac{{x}}{\mathrm{1}−{x}}}\right)\:+\:{c}\: \\ $$$$−\sqrt{{x}−{x}^{\mathrm{2}} }\:+\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\frac{{x}}{\mathrm{1}−{x}}}\right)\:+\:{c} \\ $$