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Question Number 96898 by bobhans last updated on 05/Jun/20

Commented by PRITHWISH SEN 2 last updated on 05/Jun/20

(1/6)∫{(√(x^2 +6x)) +x}dx=(1/6)∫(√((x+3)^2 −9)) dx +(1/6)∫xdx =(((x+3))/(12))(√(x^2 +6x)) −(3/4)ln∣(x+3)+(√(x^2 +6x))∣+(x^2 /(12)) +C

$$\frac{\mathrm{1}}{\mathrm{6}}\int\left\{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{6x}}\:+\mathrm{x}\right\}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{6}}\int\sqrt{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9}}\:\mathrm{dx}\:+\frac{\mathrm{1}}{\mathrm{6}}\int\mathrm{xdx} \\ $$$$=\frac{\left(\mathrm{x}+\mathrm{3}\right)}{\mathrm{12}}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{6x}}\:−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}\mid\left(\mathrm{x}+\mathrm{3}\right)+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{6x}}\mid+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{12}}\:+\mathrm{C} \\ $$

Answered by john santu last updated on 05/Jun/20

I = ∫ (((√x) dx)/((√x)((√(1+(6/x)))−1))) I = ∫ (dx/((√(1+(6/x)))−1)) set z^2 =1+(6/x) then x = (6/(z^2 −1)) dx = ((−12z)/((z^2 −1)^2 )) dz I = ∫ (1/(1−z)) . (((−12z)/((z^2 −1)^2 )) dz I = ∫ ((12z dz)/((z−1)(z^2 −1)^2 )) = ∫ ((12z dz)/((z−1)^2 (z+1))) I = ∫ ((3 dz)/(2(z−1))) + ∫ ((6 dz)/((z−1)^2 )) −∫(( 9 dz)/(2(z+1))) I =(3/2)ln ∣z−1∣ −(6/(z−1)) −(9/2)ln ∣z+1∣ + c I= (3/2)ln ∣(√((x+6)/x))−1∣ − ((6(√x))/(√(x+6))) −(9/2)ln ∣(√((x+6)/x))+1∣ + c

$$\mathrm{I}\:=\:\int\:\frac{\sqrt{\mathrm{x}}\:\:\mathrm{dx}}{\sqrt{\mathrm{x}}\left(\sqrt{\mathrm{1}+\frac{\mathrm{6}}{\mathrm{x}}}−\mathrm{1}\right)} \\ $$$$\mathrm{I}\:=\:\int\:\frac{\mathrm{dx}}{\sqrt{\mathrm{1}+\frac{\mathrm{6}}{\mathrm{x}}}−\mathrm{1}} \\ $$$$\mathrm{set}\:\mathrm{z}^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{6}}{\mathrm{x}}\:\mathrm{then}\:\mathrm{x}\:=\:\frac{\mathrm{6}}{\mathrm{z}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{dx}\:=\:\frac{−\mathrm{12z}}{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dz} \\ $$$$\mathrm{I}\:=\:\int\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{z}}\:.\:\frac{\left(−\mathrm{12z}\right.}{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dz}\: \\ $$$$\mathrm{I}\:=\:\int\:\frac{\mathrm{12z}\:\:\mathrm{dz}}{\left(\mathrm{z}−\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:=\:\int\:\frac{\mathrm{12z}\:\mathrm{dz}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{1}\right)} \\ $$$$\mathrm{I}\:=\:\int\:\frac{\mathrm{3}\:\mathrm{dz}}{\mathrm{2}\left(\mathrm{z}−\mathrm{1}\right)}\:+\:\int\:\frac{\mathrm{6}\:\mathrm{dz}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} }\:−\int\frac{\:\mathrm{9}\:\mathrm{dz}}{\mathrm{2}\left(\mathrm{z}+\mathrm{1}\right)} \\ $$$$\mathrm{I}\:=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{z}−\mathrm{1}\mid\:−\frac{\mathrm{6}}{\mathrm{z}−\mathrm{1}}\:−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{z}+\mathrm{1}\mid\:+\:\mathrm{c}\: \\ $$$$\mathrm{I}=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mid\sqrt{\frac{\mathrm{x}+\mathrm{6}}{\mathrm{x}}}−\mathrm{1}\mid\:−\:\frac{\mathrm{6}\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}+\mathrm{6}}}\:−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{ln}\:\mid\sqrt{\frac{\mathrm{x}+\mathrm{6}}{\mathrm{x}}}+\mathrm{1}\mid\:+\:\mathrm{c}\: \\ $$

Answered by Sourav mridha last updated on 05/Jun/20

=(1/6)[∫(√(x^2 +6x)) +(1/2)x^2 ] =(1/6)[∫(√((x+3)^2 −3^2 ))d(x+3)]+(1/(12))x^2 =(1/6)[(((x+3))/2).(√(x^2 +6x)) −(9/2)ln[(x+3)+(√(x^2 +6x))] +(1/(12))x^2 +c

$$=\frac{\mathrm{1}}{\mathrm{6}}\left[\int\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}}\:+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[\int\sqrt{\left(\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }\boldsymbol{{d}}\left(\boldsymbol{{x}}+\mathrm{3}\right)\right]+\frac{\mathrm{1}}{\mathrm{12}}\boldsymbol{{x}}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[\frac{\left(\boldsymbol{{x}}+\mathrm{3}\right)}{\mathrm{2}}.\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}}\:\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{9}}{\mathrm{2}}\boldsymbol{{ln}}\left[\left(\boldsymbol{{x}}+\mathrm{3}\right)+\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{12}}\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{c}} \\ $$

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