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Question Number 152270 by peter frank last updated on 27/Aug/21

∫((5x+3)/( (√(x^2 +4x+10))))dx

$$\int\frac{\mathrm{5x}+\mathrm{3}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{10}}}\mathrm{dx} \\ $$

Answered by Olaf_Thorendsen last updated on 27/Aug/21

F(x) = ∫((5x+3)/( (√(x^2 +4x+10)))) dx F(u−2) = ∫((5u−7)/( (√(u^2 +6)))) du F(u−2) = 5(√(u^2 +6))−7arcsin((u/( (√6))))+C F(x) = 5(√(x^2 +4x+10))−7arcsin(((x+2)/( (√6))))+C

$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{5}{x}+\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}\:{dx} \\ $$$$\mathrm{F}\left({u}−\mathrm{2}\right)\:=\:\int\frac{\mathrm{5}{u}−\mathrm{7}}{\:\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}}\:{du} \\ $$$$\mathrm{F}\left({u}−\mathrm{2}\right)\:=\:\mathrm{5}\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}−\mathrm{7arcsin}\left(\frac{{u}}{\:\sqrt{\mathrm{6}}}\right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7arcsin}\left(\frac{{x}+\mathrm{2}}{\:\sqrt{\mathrm{6}}}\right)+\mathrm{C} \\ $$$$ \\ $$

Commented by peter frank last updated on 27/Aug/21

sorry sir step two explanation please

$$\mathrm{sorry}\:\mathrm{sir}\:\mathrm{step}\:\mathrm{two}\:\mathrm{explanation}\: \\ $$$$\mathrm{please} \\ $$

Answered by puissant last updated on 27/Aug/21

=5∫((x+(3/5))/( (√(x^2 +4x+10))))dx = (5/2)∫((2x+(6/5))/( (√(x^2 +4x+10))))dx =(5/2)∫((2x+4−((14)/5))/( (√(x^2 +4x+10))))dx =(5/2)∫((2x+4)/( (√(x^2 +4x+10))))dx−7∫(dx/( (√(x^2 +4x+10)))) =5∫((2x+4)/(2(√(x^2 +4x+10))))dx−7∫(dx/( (√((x+2)^2 +((√6))^2 )))) =5(√(x^2 +4x+10))−7ln∣x+2+(√(x^2 +4x+10))∣+C because ∫(dx/( (√(x^2 +a^2 ))))=ln∣x+(√(x^2 +a^2 ))∣+C ∴∵ Q=5(√(x^2 +4x+10))−7ln∣x+2+(√(x^2 +4x+10))∣+C..

$$=\mathrm{5}\int\frac{{x}+\frac{\mathrm{3}}{\mathrm{5}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx}\:=\:\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\frac{\mathrm{6}}{\mathrm{5}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{4}−\frac{\mathrm{14}}{\mathrm{5}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx}−\mathrm{7}\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}} \\ $$$$=\mathrm{5}\int\frac{\mathrm{2}{x}+\mathrm{4}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx}−\mathrm{7}\int\frac{{dx}}{\:\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{6}}\right)^{\mathrm{2}} }} \\ $$$$=\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7}{ln}\mid{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}\mid+{C} \\ $$$${because}\:\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}={ln}\mid{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\mid+{C} \\ $$$$ \\ $$$$\therefore\because\:{Q}=\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7}{ln}\mid{x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}\mid+{C}.. \\ $$

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