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Question Number 18024 by mondodotto@gmail.com last updated on 13/Jul/17

Answered by sma3l2996 last updated on 14/Jul/17

I=∫((5x+3)/(√(x^2 +4x+10)))dx=(5/2)∫((2x)/(√(x^2 +4x+10)))dx+∫((3dx)/(√(x^2 +4x+10))) =(5/2)∫((2x+4)/(√(x^2 +4x+10)))+∫((3−10)/(√((x+2)^2 +6)))dx =5(√(x^2 +4x+10))−∫(7/((√6)(√((((x+2)/(√6)))^2 +1))))dx+c let t=((x+2)/(√6))⇒dt=(dx/(√6)) (7/(√6))∫(dx/(√((((x+2)/(√6)))^2 +1)))=7∫(dt/(√(t^2 +1)))=7ln(t+(√(t^2 +1)))+a I=5(√(x^2 +4x+10))−7ln(((x+2)/(√6))+(√((((x+2)/(√6)))^2 +1)))+C_1 =5(√(x^2 +4x+10))−7ln(x+2+(√(x^2 +4x+10)))−7ln(√6)+C_1 I=5(√(x^2 +4x+10))−7ln(x+2+(√(x^2 +4x+10)))+C

$${I}=\int\frac{\mathrm{5}{x}+\mathrm{3}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx}=\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}{dx}+\int\frac{\mathrm{3}{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{4}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}}+\int\frac{\mathrm{3}−\mathrm{10}}{\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{6}}}{dx} \\ $$$$=\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\int\frac{\mathrm{7}}{\sqrt{\mathrm{6}}\sqrt{\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{6}}}\right)^{\mathrm{2}} +\mathrm{1}}}{dx}+{c} \\ $$$${let}\:\:{t}=\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{6}}}\Rightarrow{dt}=\frac{{dx}}{\sqrt{\mathrm{6}}} \\ $$$$\frac{\mathrm{7}}{\sqrt{\mathrm{6}}}\int\frac{{dx}}{\sqrt{\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{6}}}\right)^{\mathrm{2}} +\mathrm{1}}}=\mathrm{7}\int\frac{{dt}}{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}=\mathrm{7}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)+{a} \\ $$$${I}=\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7}{ln}\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{6}}}+\sqrt{\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{6}}}\right)^{\mathrm{2}} +\mathrm{1}}\right)+{C}_{\mathrm{1}} \\ $$$$=\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7}{ln}\left({x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}\right)−\mathrm{7}{ln}\sqrt{\mathrm{6}}+{C}_{\mathrm{1}} \\ $$$${I}=\mathrm{5}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}−\mathrm{7}{ln}\left({x}+\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}\right)+{C} \\ $$

Commented by mondodotto@gmail.com last updated on 14/Jul/17

i preciate sir thanx

$$\mathrm{i}\:\mathrm{preciate}\:\mathrm{sir}\:\mathrm{thanx} \\ $$

Answered by Abbas-Nahi last updated on 14/Jul/17

∫(( 5x+3)/((√((x+2)^2 +6 )) )) dx let u=x+2 du=dx ∫((5(u−2)+3)/((√(u^2 +6)) )) du=∫(( 5u−7)/((√(u^2 +6)) )) du 5∫(u/((√(u^2 +6)) )) du−7∫(1/((√(u^2 +6)) ))du (5/2)∫((2u)/((√(u^2 +6)) )) du−(7/((√6) )) ∫(1/( (√(1+((u/(√6)) )^(2 ) )))) du then complete the solution....

$$\int\frac{\:\mathrm{5}{x}+\mathrm{3}}{\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{6}\:\:}\:}\:{dx} \\ $$$${let}\:{u}={x}+\mathrm{2}\:\:\:\:{du}={dx} \\ $$$$\int\frac{\mathrm{5}\left({u}−\mathrm{2}\right)+\mathrm{3}}{\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}\:}\:{du}=\int\frac{\:\mathrm{5}{u}−\mathrm{7}}{\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}\:}\:{du} \\ $$$$\mathrm{5}\int\frac{{u}}{\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}\:}\:{du}−\mathrm{7}\int\frac{\mathrm{1}}{\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}\:}{du} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}{u}}{\sqrt{{u}^{\mathrm{2}} +\mathrm{6}}\:}\:{du}−\frac{\mathrm{7}}{\sqrt{\mathrm{6}}\:}\:\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\frac{{u}}{\sqrt{\mathrm{6}}}\:\right)^{\mathrm{2}\:} \:\:\:}}\:{du} \\ $$$${then}\:{complete}\:{the}\:{solution}.... \\ $$

Commented by mondodotto@gmail.com last updated on 14/Jul/17

waoohh nice1 thanx

$$\mathrm{waoohh}\:\mathrm{nice1}\:\mathrm{thanx} \\ $$

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