Question Number 191009 by pascal889 last updated on 16/Apr/23
Answered by Frix last updated on 16/Apr/23
$$\int\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{dt}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}\int{dt}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{t}+{C} \\ $$
Commented by ARUNG_Brandon_MBU last updated on 16/Apr/23
��
Answered by Frix last updated on 16/Apr/23
$$\int\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{dx}= \\ $$$$=\int\left(\frac{\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}\left({x}+\mathrm{3}+\sqrt{\mathrm{5}}\right)}+\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}\left({x}+\mathrm{3}−\sqrt{\mathrm{5}}\right)}\right){dx}= \\ $$$$=\frac{\left(\mathrm{15}+\mathrm{7}\sqrt{\mathrm{5}}\right)\mathrm{ln}\:\mid{x}+\mathrm{3}+\sqrt{\mathrm{5}}\mid\:+\left(\mathrm{15}−\mathrm{7}\sqrt{\mathrm{5}}\right)\mathrm{ln}\:\mid{x}+\mathrm{3}−\sqrt{\mathrm{5}}\mid}{\mathrm{10}}+{C} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 16/Apr/23
$${I}=\int\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}} \\ $$$$\:\:=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{dx}−\int\frac{\mathrm{7}{dx}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}} \\ $$$$\:\:=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\mid{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}\mid−\int\frac{\mathrm{7}{dx}}{\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{5}} \\ $$$$\:\:=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\mid{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}\mid+\frac{\mathrm{7}}{\:\sqrt{\mathrm{5}}}\mathrm{argcoth}\left(\frac{{x}+\mathrm{3}}{\:\sqrt{\mathrm{5}}}\right)+{C} \\ $$$$\:\:=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\mid{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}\mid+\frac{\mathrm{7}\sqrt{\mathrm{5}}}{\mathrm{10}}\mathrm{ln}\mid\frac{{x}+\mathrm{3}+\sqrt{\mathrm{5}}}{{x}+\mathrm{3}−\sqrt{\mathrm{5}}}\mid+{C} \\ $$