x-2-x-1-3-dx-

Question Number 85952 by jagoll last updated on 26/Mar/20

$$\int\:\:\frac{\sqrt{\mathrm{x}}}{\mathrm{2}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}}}\:\mathrm{dx}\: \\ $$

Answered by john santu last updated on 26/Mar/20

let t^6 = x ⇒dx = 6t^5 dt ∫ (t^3 /(2+t^2 )) × 6t^5 dt = ∫ ((6t^8 )/(2+t^2 )) dt [ it sholud be easy to solve ]

$${let}\:{t}^{\mathrm{6}} \:=\:{x}\:\Rightarrow{dx}\:=\:\mathrm{6}{t}^{\mathrm{5}} \:{dt} \\ $$$$\int\:\:\frac{{t}^{\mathrm{3}} }{\mathrm{2}+{t}^{\mathrm{2}} }\:×\:\mathrm{6}{t}^{\mathrm{5}} \:{dt}\: \\ $$$$=\:\int\:\frac{\mathrm{6}{t}^{\mathrm{8}} }{\mathrm{2}+{t}^{\mathrm{2}} }\:{dt}\: \\ $$$$\left[\:{it}\:{sholud}\:{be}\:{easy}\:{to}\:{solve}\:\right]\: \\ $$

Commented by TANMAY PANACEA. last updated on 26/Mar/20

6∫((t^8 −16+16)/(t^2 +2)) 6∫(((t^4 +4)(t^2 +2)(t^2 −2)+16)/(t^2 +2)) 6∫t^6 −2t^4 +4t^2 −8 dt+96∫(dt/(t^2 +2)) 6((t^7 /7)−((2t^5 )/5)+((4t^3 )/3)−8t)+96×(1/( (√2)))tan^(−1) ((t/( (√2))))+c

$$\mathrm{6}\int\frac{{t}^{\mathrm{8}} −\mathrm{16}+\mathrm{16}}{{t}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\mathrm{6}\int\frac{\left({t}^{\mathrm{4}} +\mathrm{4}\right)\left({t}^{\mathrm{2}} +\mathrm{2}\right)\left({t}^{\mathrm{2}} −\mathrm{2}\right)+\mathrm{16}}{{t}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\mathrm{6}\int{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} −\mathrm{8}\:\:{dt}+\mathrm{96}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\mathrm{6}\left(\frac{{t}^{\mathrm{7}} }{\mathrm{7}}−\frac{\mathrm{2}{t}^{\mathrm{5}} }{\mathrm{5}}+\frac{\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{8}{t}\right)+\mathrm{96}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$

Commented by jagoll last updated on 26/Mar/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

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