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du-u-u-2-




Question Number 83893 by sahnaz last updated on 07/Mar/20
∫(du/(u−u^2 ))
$$\int\frac{\mathrm{du}}{\mathrm{u}−\mathrm{u}^{\mathrm{2}} } \\ $$
Commented by abdomathmax last updated on 07/Mar/20
∫  (du/(u−u^2 )) =∫  (du/(u(1−u))) =∫  ((1/u)+(1/(1−u)))du  =ln∣(u/(1−u))∣ +C
$$\int\:\:\frac{{du}}{{u}−{u}^{\mathrm{2}} }\:=\int\:\:\frac{{du}}{{u}\left(\mathrm{1}−{u}\right)}\:=\int\:\:\left(\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right){du} \\ $$$$={ln}\mid\frac{{u}}{\mathrm{1}−{u}}\mid\:+{C} \\ $$

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