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Question Number 67070 by mRDv143 last updated on 22/Aug/19

Commented by Prithwish sen last updated on 23/Aug/19

∫((x^2 −1)/((x+1)^2 (√(x^3 +x^2 +x)))) dx=∫((x^2 −1)/((x+1)^2 .x (√(x+(1/x)+1)))) dx =∫(((1−(1/x^2 )) dx)/((x+(1/x)+2)(√(x+(1/x)+1)))) putting x+(1/x) = u = ∫(du/((u+2)(√(u+1)))) putting u+1 = t^2 =2∫(dt/(t^2 +1)) = 2tan^(−1) t +C = 2 tan^(−1) (√(((x^2 +x+1)/x) )) +C please check.

$$\int\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:\sqrt{\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}}}\:\mathrm{dx}=\int\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} .\mathrm{x}\:\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{1}}}\:\mathrm{dx} \\ $$$$=\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\:\:\mathrm{dx}}{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{2}\right)\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{1}}}\:\:\mathrm{putting}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{u} \\ $$$$=\:\int\frac{\mathrm{du}}{\left(\mathrm{u}+\mathrm{2}\right)\sqrt{\mathrm{u}+\mathrm{1}}}\:\:\mathrm{putting}\:\mathrm{u}+\mathrm{1}\:=\:\mathrm{t}^{\mathrm{2}} \\ $$$$=\mathrm{2}\int\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{2}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \boldsymbol{\mathrm{t}}\:+\boldsymbol{\mathrm{C}}\:=\:\mathrm{2}\:\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \sqrt{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:}\:+\boldsymbol{\mathrm{C}} \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$

Commented by mathmax by abdo last updated on 22/Aug/19

i see a small error if we put x+(1/x)=u ⇒ ∫ ((x^2 −1)/((x+1)^2 (√(x^3 +x^2 +x))))dx =∫ (du/((u+2)(√(u+1)))) =_(u+1=t^2 ) ∫((2tdt)/((t^2 +1)t)) =2 ∫ (dt/(1+t^2 )) =2arctan(t)+c =2arctan((√(u+1))) +c =2arctan((√(x+(1/x)+1)))+c

$${i}\:{see}\:{a}\:{small}\:{error}\:{if}\:{we}\:{put}\:{x}+\frac{\mathrm{1}}{{x}}={u}\:\Rightarrow \\ $$$$\int\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} \:+{x}}}{dx}\:=\int\:\:\:\frac{{du}}{\left({u}+\mathrm{2}\right)\sqrt{{u}+\mathrm{1}}}\:\:=_{{u}+\mathrm{1}={t}^{\mathrm{2}} } \:\int\frac{\mathrm{2}{tdt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right){t}} \\ $$$$=\mathrm{2}\:\int\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{2}{arctan}\left({t}\right)+{c}\:=\mathrm{2}{arctan}\left(\sqrt{{u}+\mathrm{1}}\right)\:+{c} \\ $$$$=\mathrm{2}{arctan}\left(\sqrt{{x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}}\right)+{c}\: \\ $$

Commented by Prithwish sen last updated on 23/Aug/19

Thank you very much sir.It was a typo. I correct it.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}.\mathrm{It}\:\mathrm{was}\:\mathrm{a}\:\mathrm{typo}.\:\mathrm{I}\: \\ $$$$\mathrm{correct}\:\mathrm{it}. \\ $$

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