Question-123967

Question Number 123967 by Algoritm last updated on 29/Nov/20

Answered by TANMAY PANACEA last updated on 29/Nov/20

t^6 =x+2 ∫(t/(t^2 +t^3 ))×6t^5 dt ∫((6t^4 )/(1+t))dt 6∫((t^4 −1+1)/(t+1))dt 6∫(((t^2 +1)(t+1)(t−1)+1)/(t+1))dt 6∫t^3 −t^2 +t−1+(1/(t+1)) dt 6{(t^4 /4)−(t^3 /3)+(t^2 /2)−t+ln(t+1)}+c put t=(x+2)^(1/6)

$${t}^{\mathrm{6}} ={x}+\mathrm{2} \\ $$$$\int\frac{{t}}{{t}^{\mathrm{2}} +{t}^{\mathrm{3}} }×\mathrm{6}{t}^{\mathrm{5}} {dt} \\ $$$$\int\frac{\mathrm{6}{t}^{\mathrm{4}} }{\mathrm{1}+{t}}{dt} \\ $$$$\mathrm{6}\int\frac{{t}^{\mathrm{4}} −\mathrm{1}+\mathrm{1}}{{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{6}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)+\mathrm{1}}{{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{6}\int{t}^{\mathrm{3}} −{t}^{\mathrm{2}} +{t}−\mathrm{1}+\frac{\mathrm{1}}{{t}+\mathrm{1}}\:\:{dt} \\ $$$$\mathrm{6}\left\{\frac{{t}^{\mathrm{4}} }{\mathrm{4}}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−{t}+{ln}\left({t}+\mathrm{1}\right)\right\}+{c} \\ $$$${put}\:{t}=\left({x}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$

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Question-123967

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