Question Number 152270 by peter frank last updated on 27/Aug/21
$$\int\frac{\mathrm{5x}+\mathrm{3}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{10}}}\mathrm{dx} \\ $$
Answered by Olaf_Thorendsen last updated on 27/Aug/21
$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{5}{x}+\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}\:{dx} \\ $$$$\mathrm{F}\left({u}−\mathrm{2}\right)\:=\:\int\frac{\mathrm{5}{u}−\mathrm{7}}{\:\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}}\:{du} \\ $$$$\mathrm{F}\left({u}−\mathrm{2}\right)\:=\:\mathrm{5}\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}−\mathrm{7arcsin}\left(\frac{{u}}{\:\sqrt{\mathrm{6}}}\right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7arcsin}\left(\frac{{x}+\mathrm{2}}{\:\sqrt{\mathrm{6}}}\right)+\mathrm{C} \\ $$$$ \\ $$
Commented by peter frank last updated on 27/Aug/21
$$\mathrm{sorry}\:\mathrm{sir}\:\mathrm{step}\:\mathrm{two}\:\mathrm{explanation}\: \\ $$$$\mathrm{please} \\ $$
Answered by puissant last updated on 27/Aug/21
$$=\mathrm{5}\int\frac{{x}+\frac{\mathrm{3}}{\mathrm{5}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx}\:=\:\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\frac{\mathrm{6}}{\mathrm{5}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{4}−\frac{\mathrm{14}}{\mathrm{5}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx}−\mathrm{7}\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}} \\ $$$$=\mathrm{5}\int\frac{\mathrm{2}{x}+\mathrm{4}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx}−\mathrm{7}\int\frac{{dx}}{\:\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{6}}\right)^{\mathrm{2}} }} \\ $$$$=\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7}{ln}\mid{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}\mid+{C} \\ $$$${because}\:\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}={ln}\mid{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\mid+{C} \\ $$$$ \\ $$$$\therefore\because\:{Q}=\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7}{ln}\mid{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}\mid+{C}.. \\ $$