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Question Number 177193 by peter frank last updated on 02/Oct/22

∫ (x^(2021) /(x^(2022) +x^(4043) ))dx

$$\int\:\:\:\frac{\mathrm{x}^{\mathrm{2021}} }{\mathrm{x}^{\mathrm{2022}} +\mathrm{x}^{\mathrm{4043}} }\mathrm{dx} \\ $$

Answered by cortano1 last updated on 02/Oct/22

(1/(x(x^(2021) +1))) =(1/x) −(x^(2020) /(x^(2021) +1)) I=∫ (dx/x)−∫ (x^(2020) /(x^(2021) +1)) I= ln ∣x∣ −(1/(2021)) ∫ ((d(x^(2021) +1))/(x^(2021) +1)) I= ln ∣x∣−((ln ∣x^(2021) +1∣)/(2021)) +c I=ln ∣(x/( ((x^(2021) +1))^(1/(2021)) )) ∣ + c

$$\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2021}} +\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{x}}\:−\frac{\mathrm{x}^{\mathrm{2020}} }{\mathrm{x}^{\mathrm{2021}} +\mathrm{1}} \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{dx}}{\mathrm{x}}−\int\:\frac{\mathrm{x}^{\mathrm{2020}} }{\mathrm{x}^{\mathrm{2021}} +\mathrm{1}} \\ $$$$\mathrm{I}=\:\mathrm{ln}\:\mid\mathrm{x}\mid\:−\frac{\mathrm{1}}{\mathrm{2021}}\:\int\:\frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{2021}} +\mathrm{1}\right)}{\mathrm{x}^{\mathrm{2021}} +\mathrm{1}} \\ $$$$\mathrm{I}=\:\mathrm{ln}\:\mid\mathrm{x}\mid−\frac{\mathrm{ln}\:\mid\mathrm{x}^{\mathrm{2021}} +\mathrm{1}\mid}{\mathrm{2021}}\:+\mathrm{c}\: \\ $$$$\mathrm{I}=\mathrm{ln}\:\mid\frac{\mathrm{x}}{\:\sqrt[{\mathrm{2021}}]{\mathrm{x}^{\mathrm{2021}} +\mathrm{1}}}\:\mid\:+\:\mathrm{c}\: \\ $$

Commented by peter frank last updated on 02/Oct/22

thank you

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

Answered by som(math1967) last updated on 02/Oct/22

∫(x^(2021) /(x^(2022) (1+x^(2021) )))dx ∫(dx/(x(1+x^(2021) ))) ∫((x^(2020) dx)/(x^(2021) (1+x^(2021) ))) let x^(2021) =t⇒x^(2020) dx=(dt/(2021)) (1/(2021))∫(dt/(t(t+1))) =(1/(2021))∫(dt/t)−(1/(2021))∫(dt/(t+1)) =(1/(2021))[lnt−ln∣t+1∣] +C =(1/(2021))ln∣(x^(2021) /(x^(2021) +1))∣ +C

$$\int\frac{{x}^{\mathrm{2021}} }{{x}^{\mathrm{2022}} \left(\mathrm{1}+{x}^{\mathrm{2021}} \right)}{dx} \\ $$$$\int\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2021}} \right)} \\ $$$$\int\frac{{x}^{\mathrm{2020}} {dx}}{{x}^{\mathrm{2021}} \left(\mathrm{1}+{x}^{\mathrm{2021}} \right)} \\ $$$${let}\:{x}^{\mathrm{2021}} ={t}\Rightarrow{x}^{\mathrm{2020}} {dx}=\frac{{dt}}{\mathrm{2021}} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2021}}\int\frac{{dt}}{{t}\left({t}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2021}}\int\frac{{dt}}{{t}}−\frac{\mathrm{1}}{\mathrm{2021}}\int\frac{{dt}}{{t}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2021}}\left[{lnt}−{ln}\mid{t}+\mathrm{1}\mid\right]\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2021}}{ln}\mid\frac{{x}^{\mathrm{2021}} }{{x}^{\mathrm{2021}} +\mathrm{1}}\mid\:+{C} \\ $$$$\:\: \\ $$

Commented by peter frank last updated on 02/Oct/22

thank you

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

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