Question Number 84326 by sahnaz last updated on 11/Mar/20
$$\int\frac{\mathrm{3}−\mathrm{7u}}{\mathrm{7u}^{\mathrm{2}} −\mathrm{7}}\mathrm{du} \\ $$
Answered by TANMAY PANACEA last updated on 11/Mar/20
![(3/7)∫(du/((u+1)(u−1)))−(1/2)∫((d(7u^2 −7))/(7u^2 −7)) (3/(7×2))∫(((u+1)−(u−1))/((u+1)(u−1)))du−(1/2)∫((d(7u^2 −7))/(7u^2 −7)) (3/(14))[∫(du/(u−1))−∫(du/(u+1))]−(1/2)∫((d(7u^2 −7))/(7u^2 −7)) (3/(14))[ln(((u−1)/(u+1)))]−(1/2)ln(7u^2 −7)+c (3/(14))ln(((u−1)/(u+1)))−(1/2){ln7+ln(u^2 −1)}+c (3/(14))ln(((u−1)/(u+1)))−(1/2)ln(u^2 −1)+C_1](Q84332.png)
$$\frac{\mathrm{3}}{\mathrm{7}}\int\frac{{du}}{\left({u}+\mathrm{1}\right)\left({u}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left(\mathrm{7}{u}^{\mathrm{2}} −\mathrm{7}\right)}{\mathrm{7}{u}^{\mathrm{2}} −\mathrm{7}} \\ $$$$\frac{\mathrm{3}}{\mathrm{7}×\mathrm{2}}\int\frac{\left({u}+\mathrm{1}\right)−\left({u}−\mathrm{1}\right)}{\left({u}+\mathrm{1}\right)\left({u}−\mathrm{1}\right)}{du}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left(\mathrm{7}{u}^{\mathrm{2}} −\mathrm{7}\right)}{\mathrm{7}{u}^{\mathrm{2}} −\mathrm{7}} \\ $$$$\frac{\mathrm{3}}{\mathrm{14}}\left[\int\frac{{du}}{{u}−\mathrm{1}}−\int\frac{{du}}{{u}+\mathrm{1}}\right]−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left(\mathrm{7}{u}^{\mathrm{2}} −\mathrm{7}\right)}{\mathrm{7}{u}^{\mathrm{2}} −\mathrm{7}} \\ $$$$\frac{\mathrm{3}}{\mathrm{14}}\left[{ln}\left(\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\right)\right]−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{7}{u}^{\mathrm{2}} −\mathrm{7}\right)+{c} \\ $$$$\frac{\mathrm{3}}{\mathrm{14}}{ln}\left(\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\mathrm{7}+{ln}\left({u}^{\mathrm{2}} −\mathrm{1}\right)\right\}+{c} \\ $$$$\frac{\mathrm{3}}{\mathrm{14}}{ln}\left(\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}^{\mathrm{2}} −\mathrm{1}\right)+{C}_{\mathrm{1}} \\ $$