Question Number 84327 by sahnaz last updated on 11/Mar/20
$$\int\frac{\mathrm{3}−\mathrm{7u}}{\mathrm{7u}^{\mathrm{2}} −\mathrm{7}}\mathrm{du} \\ $$
Answered by 20092104 last updated on 15/Mar/20
$$\int\frac{\mathrm{3}−\mathrm{7}{u}}{\mathrm{7}{u}^{\mathrm{2}} −\mathrm{7}}{du} \\ $$$$=\int\frac{\mathrm{3}−\mathrm{7}{u}}{\mathrm{7}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}{du} \\ $$$$=\int\left(\frac{\mathrm{3}}{\mathrm{7}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}−\frac{{u}}{{u}^{\mathrm{2}} −\mathrm{1}}\right){du} \\ $$$$=\frac{\mathrm{3}}{\mathrm{7}}\int\frac{\mathrm{1}}{\left({u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)}{du}−\int\frac{{u}}{{u}^{\mathrm{2}} −\mathrm{1}}{du} \\ $$$$=\frac{\mathrm{3}}{\mathrm{7}}\int\left(\frac{\mathrm{1}}{\mathrm{2}\left({u}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({u}+\mathrm{1}\right)}\right){du}−\int\frac{{u}}{{u}^{\mathrm{2}} −\mathrm{1}}{du} \\ $$$$=\frac{\mathrm{3}}{\mathrm{14}}\left({ln}\mid{u}−\mathrm{1}\mid+{ln}\mid{u}+\mathrm{1}\mid\right)−\int\frac{{u}}{{u}^{\mathrm{2}} −\mathrm{1}}{du} \\ $$$${let}\:{v}={u}^{\mathrm{2}} −\mathrm{1},\:{dv}=\mathrm{2}{udu},\:{du}=\frac{{dv}}{\mathrm{2}{u}} \\ $$$$\frac{\mathrm{3}}{\mathrm{14}}{ln}\mid{u}−\mathrm{1}\mid+\frac{\mathrm{3}}{\mathrm{14}}{ln}\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{v}}{dv} \\ $$$$=\frac{\mathrm{3}}{\mathrm{14}}{ln}\mid{u}−\mathrm{1}\mid+\frac{\mathrm{3}}{\mathrm{14}}{ln}\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{u}^{\mathrm{2}} −\mathrm{1}\mid+{C} \\ $$