Question Number 86708 by jagoll last updated on 31/Mar/20
$$\int\mathrm{x}\:\sqrt{\sqrt{\mathrm{2}}\:\mathrm{x}−\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx}\: \\ $$
Answered by john santu last updated on 31/Mar/20
![let : (√(2 ))x −(√(2x^2 −1)) = u (i) ⇒ ((2x^2 −(2x^2 −1))/((√2) x^2 +(√(2x^2 −1)))) = u ⇒ (1/((√2) x^2 +(√(2x^2 −1)))) = u (√2) x^2 + (√(2x^2 −1)) = (1/u) (ii) (i)+(ii) ⇒2(√2) x^2 = u+(1/u) 4(√2) xdx = 1−(1/u^2 ) du x dx = (1/(4(√2) ))(1−(1/u^2 )) du I = (1/(4(√2)))∫ (1−(1/u^2 )) u^(1/2) du I = ((√2)/8) ∫ (u^(1/2) −u^(−(3/2)) ) du I = ((√2)/8) [ (2/3)u^(3/2) + (2/(√u)) ] + c I = ((√2)/(12)) ((√2) x−(√(2x^2 −1)))^(3/2) + ((√2)/(4(√((√2) x−(√(2x^2 −1)))))) + c](Q86711.png)
$$\mathrm{let}\::\:\sqrt{\mathrm{2}\:}\mathrm{x}\:−\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}\:=\:\mathrm{u}\:\left(\mathrm{i}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{2x}^{\mathrm{2}} −\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\right)}{\sqrt{\mathrm{2}}\:\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}}\:=\:\mathrm{u} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}}\:=\:\mathrm{u} \\ $$$$\sqrt{\mathrm{2}}\:\mathrm{x}^{\mathrm{2}} \:+\:\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{u}}\:\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\:\Rightarrow\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{u}+\frac{\mathrm{1}}{\mathrm{u}} \\ $$$$\mathrm{4}\sqrt{\mathrm{2}}\:\mathrm{xdx}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }\:\mathrm{du}\: \\ $$$$\mathrm{x}\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}\:}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }\right)\:\mathrm{du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\int\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }\right)\:\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{du}\: \\ $$$$\mathrm{I}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\:\int\:\left(\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{u}^{−\frac{\mathrm{3}}{\mathrm{2}}} \right)\:\mathrm{du}\: \\ $$$$\mathrm{I}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\:\left[\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}^{\frac{\mathrm{3}}{\mathrm{2}}} +\:\frac{\mathrm{2}}{\sqrt{\mathrm{u}}}\:\right]\:+\:\mathrm{c} \\ $$$$\mathrm{I}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{12}}\:\left(\sqrt{\mathrm{2}}\:\mathrm{x}−\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\: \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{\sqrt{\mathrm{2}}\:\mathrm{x}−\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}}}\:+\:\mathrm{c}\: \\ $$
Commented by jagoll last updated on 30/Mar/20
$$\mathrm{wow}====\mathrm{easy}\:\mathrm{cool}\:\mathrm{sir} \\ $$$$\mathrm{thank}\:\mathrm{you} \\ $$$$ \\ $$
Commented by sakeefhasan05@gmail.com last updated on 30/Mar/20
