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Question Number 35456 by math1967 last updated on 19/May/18

∫(dx/(x(x^(2018) +1)))

$$\int\frac{{dx}}{{x}\left({x}^{\mathrm{2018}} +\mathrm{1}\right)} \\ $$

Answered by ajfour last updated on 19/May/18

I=−(1/(2018))∫ ((−2018x^(−19) dx)/(1+x^(−2018) )) =−(1/(2018))ln ∣1+(1/x^(2018) )∣+c .

$${I}=−\frac{\mathrm{1}}{\mathrm{2018}}\int\:\frac{−\mathrm{2018}{x}^{−\mathrm{19}} {dx}}{\mathrm{1}+{x}^{−\mathrm{2018}} } \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{2018}}\mathrm{ln}\:\mid\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2018}} }\mid+{c}\:\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/May/18

multiply N_(r ) and D_(r ) by x^(2017) ∫((x^(2017 ) dx)/(x^(2018) (1+x^(2018) ))) t=x^(2018 ) so dt=2018×x^(2017) dx (1/(2018))∫(dt/(t(t+1))) =(1/(2018))∫(((t+1)−t)/(t(t+1))) dt =(1/(2018)){∫(dt/t)−∫(dt/(t+1))} =(1/(2018)){lnt−ln(t+1)} =(1/(2018))ln((t/(t+1))) =(1/(2018))ln((x^(2018) /(x^(2018) +1)))

$${multiply}\:{N}_{{r}\:} {and}\:{D}_{{r}\:} {by}\:{x}^{\mathrm{2017}} \\ $$$$\int\frac{{x}^{\mathrm{2017}\:} {dx}}{{x}^{\mathrm{2018}} \left(\mathrm{1}+{x}^{\mathrm{2018}} \right)} \\ $$$${t}={x}^{\mathrm{2018}\:\:\:} {so}\:{dt}=\mathrm{2018}×{x}^{\mathrm{2017}} {dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2018}}\int\frac{{dt}}{{t}\left({t}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2018}}\int\frac{\left({t}+\mathrm{1}\right)−{t}}{{t}\left({t}+\mathrm{1}\right)}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2018}}\left\{\int\frac{{dt}}{{t}}−\int\frac{{dt}}{{t}+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2018}}\left\{{lnt}−{ln}\left({t}+\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2018}}{ln}\left(\frac{{t}}{{t}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2018}}{ln}\left(\frac{{x}^{\mathrm{2018}} }{{x}^{\mathrm{2018}} +\mathrm{1}}\right) \\ $$

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