Question Number 84334 by sahnaz last updated on 11/Mar/20
$$\int\frac{\mathrm{1}−\mathrm{u}}{−\mathrm{1}−\mathrm{2u}+\mathrm{u}^{\mathrm{2}} }\mathrm{du} \\ $$
Commented by niroj last updated on 11/Mar/20
$$\:\:\int\:\frac{−\left(\mathrm{u}−\mathrm{1}\right)}{−\left(\mathrm{1}+\mathrm{2u}−\mathrm{u}^{\mathrm{2}} \right)}\mathrm{du} \\ $$$$\:\:\:\int\:\frac{{u}−\mathrm{1}}{\mathrm{1}+\mathrm{2}{u}−{u}^{\mathrm{2}} }{du} \\ $$$$\:\:\:\int\:\frac{{u}−\mathrm{1}}{\mathrm{1}+\mathrm{2}{u}−{u}^{\mathrm{2}} }{du} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}\left({u}−\mathrm{1}\right)}{\mathrm{1}+\mathrm{2}{u}−{u}^{\mathrm{2}} }{du} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{1}+\mathrm{2}{u}−{u}^{\mathrm{2}} \right)+{c} \\ $$
Answered by MJS last updated on 11/Mar/20
$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}−\mathrm{1}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}−\mathrm{1}+\sqrt{\mathrm{2}}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({u}−\mathrm{1}−\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({u}−\mathrm{1}+\sqrt{\mathrm{2}}\right)/= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({u}^{\mathrm{2}} −\mathrm{2}{u}−\mathrm{1}\right)\:+{C} \\ $$