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Question Number 57433 by naka3546 last updated on 04/Apr/19

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

∫((x^4 +1)/(x^2 (√(x^2 (x^2 −(1/x^2 ))))))dx ∫((x+(1/x^3 ))/(√(x^2 −(1/x^2 ))))dx t^2 =x^2 −(1/x^2 ) 2tdt=(2x+(2/x^3 ))dx tdt=(x+(1/x^3 ))dx ∫((tdt)/t) =t+c =(√(x^2 −(1/x^2 ) )) +c

$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}}{dx} \\ $$$$\int\frac{{x}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}{\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{dx} \\ $$$${t}^{\mathrm{2}} ={x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}{tdt}=\left(\mathrm{2}{x}+\frac{\mathrm{2}}{{x}^{\mathrm{3}} }\right){dx} \\ $$$${tdt}=\left({x}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right){dx} \\ $$$$\int\frac{{tdt}}{{t}} \\ $$$$={t}+{c} \\ $$$$=\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:}\:\:+{c} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19

pls check ...tricky problem...

$${pls}\:{check}\:...{tricky}\:{problem}... \\ $$

Commented by MJS last updated on 04/Apr/19

your solution is correct

$$\mathrm{your}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{correct} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Apr/19

thank you sir...

$${thank}\:{you}\:{sir}... \\ $$

Commented by malwaan last updated on 05/Apr/19

and (√(x^2 −(1/x^2 ))) =((√(x^4 −1))/x) thank you so much

$$\boldsymbol{{and}}\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }}\:=\frac{\sqrt{\boldsymbol{{x}}^{\mathrm{4}} −\mathrm{1}}}{\boldsymbol{{x}}}\: \\ $$$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{so}}\:\boldsymbol{{much}} \\ $$

Answered by MJS last updated on 04/Apr/19

possible hat trick: ∫((x^4 +1)/(x^2 (√(x^4 −1))))dx=∫((2x^2 )/(√(x^4 −1)))dx−∫((√(x^4 −1))/x^2 )dx now write ∫((√(x^4 −1))/x^2 )dx by parts without solving: ∫u′v=uv−∫uv′ u′=(1/x^2 ) ⇒ u=−(1/x) v=(√(x^4 −1)) ⇒ v′=((2x^3 )/(√(x^4 −1))) so we have: ∫((x^4 +1)/(x^2 (√(x^4 −1))))dx=∫((2x^2 )/(√(x^4 −1)))dx−(−((√(x^4 −1))/x)−∫−((2x^2 )/(√(x^4 −1)))dx)= =((√(x^4 −1))/x)+C

$$\mathrm{possible}\:\mathrm{hat}\:\mathrm{trick}: \\ $$$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}=\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}−\int\frac{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}^{\mathrm{2}} }{dx} \\ $$$$\mathrm{now}\:\mathrm{write}\:\int\frac{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}^{\mathrm{2}} }{dx}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{without} \\ $$$$\mathrm{solving}: \\ $$$$\int{u}'{v}={uv}−\int{uv}' \\ $$$${u}'=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:{u}=−\frac{\mathrm{1}}{{x}} \\ $$$${v}=\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}\:\Rightarrow\:{v}'=\frac{\mathrm{2}{x}^{\mathrm{3}} }{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}=\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}−\left(−\frac{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}}−\int−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}\right)= \\ $$$$=\frac{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}}+{C} \\ $$

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