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Question Number 36597 by rahul 19 last updated on 03/Jun/18

∫(dx/(x^(2/3) (1+x^(2/3) ))) = ?

$$\int\frac{\mathrm{d}{x}}{{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}+{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)}\:=\:? \\ $$

Commented by rahul 19 last updated on 03/Jun/18

I tried by taking x^(2/3) common from the bracket of denominator and then (x^((−2)/3) +1)=t but couldn′t reach till end. Can someone try like this?

$$\mathrm{I}\:\mathrm{tried}\:\mathrm{by}\:\mathrm{taking}\:{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \:{common}\:{from} \\ $$$${the}\:{bracket}\:{of}\:{denominator}\:{and}\: \\ $$$${then}\:\left({x}^{\frac{−\mathrm{2}}{\mathrm{3}}} +\mathrm{1}\right)={t}\:{but}\:{couldn}'\mathrm{t}\:\mathrm{reach}\:\mathrm{till} \\ $$$$\mathrm{end}.\:\mathrm{Can}\:\mathrm{someone}\:\mathrm{try}\:\mathrm{like}\:\mathrm{this}? \\ $$

Commented by abdo mathsup 649 cc last updated on 03/Jun/18

changement x^(2/3) =t give x=t^(3/2) I = ∫ (1/(t( 1+t))) (3/2) t^(1/2) dt = (3/2) ∫ ((√t)/(t(1+t)))dt and changement (√t) =x give I = (3/2) ∫ (x/(x^2 ( 1+x^2 ))) 2x dx = 3 ∫ (dx/(1+x^2 )) = 3 arctanx +c =3 arctan(t(√t)) +c .

$${changement}\:{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \:\:={t}\:{give}\:{x}={t}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${I}\:\:=\:\int\:\:\:\frac{\mathrm{1}}{{t}\left(\:\mathrm{1}+{t}\right)}\:\frac{\mathrm{3}}{\mathrm{2}}\:{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:{dt} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}\:\int\:\:\:\frac{\sqrt{{t}}}{{t}\left(\mathrm{1}+{t}\right)}{dt}\:\:{and}\:{changement}\:\sqrt{{t}}\:\:={x}\:{give} \\ $$$${I}\:\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\int\:\:\:\frac{{x}}{{x}^{\mathrm{2}} \left(\:\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\mathrm{2}{x}\:{dx} \\ $$$$=\:\mathrm{3}\:\int\:\:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\mathrm{3}\:{arctanx}\:+{c} \\ $$$$=\mathrm{3}\:{arctan}\left({t}\sqrt{{t}}\right)\:+{c}\:. \\ $$$$ \\ $$

Answered by Joel579 last updated on 03/Jun/18

I = ∫ (dx/(x^(2/3) (1 + x^(2/3) ))) (t = x^(1/3) → dt = (x^(−2/3) /3) dx ⇔ dx = 3t^2 dt) = ∫ ((3t^2 )/(t^2 (1 + t^2 ))) dt = 3 tan^(−1) (t) + C = 3 tan^(−1) (x^(1/3) ) + C

$${I}\:=\:\int\:\frac{{dx}}{{x}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{1}\:+\:{x}^{\mathrm{2}/\mathrm{3}} \right)}\:\:\:\left({t}\:=\:{x}^{\mathrm{1}/\mathrm{3}} \:\:\rightarrow\:\:{dt}\:=\:\frac{{x}^{−\mathrm{2}/\mathrm{3}} }{\mathrm{3}}\:{dx}\:\:\Leftrightarrow\:\:{dx}\:=\:\mathrm{3}{t}^{\mathrm{2}} \:{dt}\right) \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{3}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \left(\mathrm{1}\:+\:{t}^{\mathrm{2}} \right)}\:{dt} \\ $$$$\:\:\:=\:\mathrm{3}\:\mathrm{tan}^{−\mathrm{1}} \left({t}\right)\:+\:{C} \\ $$$$\:\:\:=\:\mathrm{3}\:\mathrm{tan}^{−\mathrm{1}} \left({x}^{\mathrm{1}/\mathrm{3}} \right)\:+\:{C} \\ $$

Answered by sma3l2996 last updated on 03/Jun/18

let t=x^(1/3) ⇒dt=(dx/(3x^(2/3) )) x^(2/3) =t^2 ∫(dx/(x^(2/3) (1+x^(2/3) )))=3∫(dt/(1+t^2 ))=3tan^(−1) (t)+C =3tan^(−1) (x^(1/3) )+C

$${let}\:\:{t}={x}^{\mathrm{1}/\mathrm{3}} \Rightarrow{dt}=\frac{{dx}}{\mathrm{3}{x}^{\mathrm{2}/\mathrm{3}} } \\ $$$${x}^{\mathrm{2}/\mathrm{3}} ={t}^{\mathrm{2}} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{2}/\mathrm{3}} \right)}=\mathrm{3}\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{3}{tan}^{−\mathrm{1}} \left({t}\right)+{C} \\ $$$$=\mathrm{3}{tan}^{−\mathrm{1}} \left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)+{C} \\ $$

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