Question Number 213267 by MrGaster last updated on 02/Nov/24
$$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \frac{{x}−{x}^{\mathrm{2}} +{x}^{\mathrm{3}} −\ldots−{x}^{\mathrm{2018}} }{\left(\mathrm{1}+{x}\right)^{\mathrm{2021}} }{dx} \\ $$
Answered by TonyCWX08 last updated on 02/Nov/24
$${The}\:{sequence}\:{above}\:{is}\:{a}\:{geometric}\:{progression}\:{with} \\ $$$${a}={x} \\ $$$${r}=−{x} \\ $$$${n}=\mathrm{2018} \\ $$$${Hence},\:{the}\:{sum}\:{is} \\ $$$$\frac{{a}\left({r}^{{n}} −\mathrm{1}\right)}{\left({r}−\mathrm{1}\right)} \\ $$$$=\frac{{x}\left(\left(−{x}\right)^{\mathrm{2018}} −\mathrm{1}\right)}{−{x}−\mathrm{1}} \\ $$$$=\frac{{x}\left({x}^{\mathrm{2018}} −\mathrm{1}\right)}{−\left({x}+\mathrm{1}\right)} \\ $$$$ \\ $$$$\frac{\frac{{x}\left({x}^{\mathrm{2018}} −\mathrm{1}\right)}{−\left({x}+\mathrm{1}\right)}}{\left({x}+\mathrm{1}\right)^{\mathrm{2021}} } \\ $$$$=\frac{{x}\left({x}^{\mathrm{2018}} −\mathrm{1}\right)}{−\left({x}+\mathrm{1}\right)^{\mathrm{2022}} } \\ $$$$ \\ $$$${I}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{{x}\left({x}^{\mathrm{2018}} −\mathrm{1}\right)}{−\left({x}+\mathrm{1}\right)^{\mathrm{2022}} }\right){dx} \\ $$$$=−\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{x}\left({x}^{\mathrm{2018}} −\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2022}} }{dx} \\ $$$${let}\:{u}\:=\:{x}+\mathrm{1} \\ $$$${du}\:=\:{dx} \\ $$$${x}={u}−\mathrm{1} \\ $$$${x}^{\mathrm{2018}} −\mathrm{1}=\left({u}−\mathrm{1}\right)^{\mathrm{2018}} −\mathrm{1} \\ $$$${I}=−\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\left({u}−\mathrm{1}\right)\left(\left({u}−\mathrm{1}\right)^{\mathrm{2018}} −\mathrm{1}\right)}{{u}^{\mathrm{2022}} }{du} \\ $$$$=−\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\left({u}−\mathrm{1}\right)^{\mathrm{2019}} −{u}+\mathrm{1}}{{u}^{\mathrm{2022}} }{du} \\ $$$$=−\left(\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\left({u}−\mathrm{1}\right)^{\mathrm{2019}} }{{u}^{\mathrm{2022}} }{du}−\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{{u}}{{u}^{\mathrm{2022}} }{du}+\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\mathrm{1}}{{u}^{\mathrm{2022}} }{du}\right) \\ $$$$=−\left(\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\left({u}−\mathrm{1}\right)^{\mathrm{2019}} }{{u}^{\mathrm{2022}} }{du}−\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{{u}}{{u}^{\mathrm{2022}} }{du}+\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\mathrm{1}}{{u}^{\mathrm{2022}} }{du}\right) \\ $$$$=−\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\left({u}−\mathrm{1}\right)^{\mathrm{2019}} }{{u}^{\mathrm{2022}} }{du}+\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\mathrm{1}}{{u}^{\mathrm{2021}} }{du}−\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\mathrm{1}}{{u}^{\mathrm{2022}} }{du} \\ $$$$=−\underset{\mathrm{1}} {\overset{\infty} {\int}}\:\frac{\left({u}−\mathrm{1}\right)^{\mathrm{2019}} }{{u}^{\mathrm{2022}} }{du}+\frac{\mathrm{1}}{\mathrm{2020}}−\frac{\mathrm{1}}{\mathrm{2021}} \\ $$$${I}'{m}\:{stuck}. \\ $$$${Can}\:{continue}? \\ $$