Question Number 85591 by sahnaz last updated on 23/Mar/20
$$\int\frac{\mathrm{1}+\mathrm{4u}}{−\mathrm{4u}^{\mathrm{2}} +\mathrm{2u}+\mathrm{2}}\mathrm{du} \\ $$$$ \\ $$
Answered by john santu last updated on 23/Mar/20
$$−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{2}{u}+\mathrm{1}\right)}{du}+\int\:\frac{\mathrm{5}}{\mathrm{3}\left({u}−\mathrm{1}\right)}{du} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\mid\mathrm{2}{u}+\mathrm{1}\mid\:+\frac{\mathrm{5}}{\mathrm{3}}\mathrm{ln}\:\mid{u}−\mathrm{1}\mid\right\}\:+{c} \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\mid\mathrm{2}{u}+\mathrm{1}\mid\:−\frac{\mathrm{5}}{\mathrm{6}}\mathrm{ln}\:\mid{u}−\mathrm{1}\mid\:+\:{c} \\ $$