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2-u-u-2-u-du-




Question Number 85319 by sahnaz last updated on 20/Mar/20
∫((2+u)/(−u^2 −u))du
$$\int\frac{\mathrm{2}+\mathrm{u}}{−\mathrm{u}^{\mathrm{2}} −\mathrm{u}}\mathrm{du} \\ $$
Answered by MJS last updated on 20/Mar/20
∫((2+u)/(−u^2 −u))du=−∫((u+2)/(u(u+1)))du=  =∫(du/(u+1))−2∫(du/u)=ln ∣u+1∣ −2ln ∣u∣ +C
$$\int\frac{\mathrm{2}+{u}}{−{u}^{\mathrm{2}} −{u}}{du}=−\int\frac{{u}+\mathrm{2}}{{u}\left({u}+\mathrm{1}\right)}{du}= \\ $$$$=\int\frac{{du}}{{u}+\mathrm{1}}−\mathrm{2}\int\frac{{du}}{{u}}=\mathrm{ln}\:\mid{u}+\mathrm{1}\mid\:−\mathrm{2ln}\:\mid{u}\mid\:+{C} \\ $$

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