Question Number 47740 by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18
$$\int\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)…\left({x}+{n}\right)} \\ $$
Answered by ajfour last updated on 14/Nov/18
$$\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{r}\right)…\left({x}+{n}\right)} \\ $$$$=\:\Sigma\frac{{A}_{{r}} }{{x}+{r}}\: \\ $$$${A}_{{r}} \:=\:\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{{r}} \:{r}!\left({n}−{r}\right)!} \\ $$$$\Rightarrow{I}\:=\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{{r}} \:{r}!\left({n}−{r}\right)!}\:\mathrm{ln}\:\left({x}+{r}\right)+{c}\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18
$${excellent}… \\ $$
Commented by ajfour last updated on 14/Nov/18
$${as}\:{example} \\ $$$$\:\:{I}\:=\:\int\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid{x}\mid−\mathrm{ln}\:\mid{x}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid{x}+\mathrm{2}\mid+{c}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18
$$\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)…\left({x}+{n}\right)}=\frac{{a}_{\mathrm{0}} }{{x}}+\frac{{a}_{\mathrm{1}} }{{x}+\mathrm{1}}+\frac{{a}_{\mathrm{2}} }{{x}+\mathrm{2}}+…+\frac{{a}_{{n}} }{{x}+{n}} \\ $$$${so} \\ $$$$\mathrm{1}={a}_{\mathrm{0}} \left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)…\left({x}+{n}\right)+{a}_{\mathrm{1}} \left({x}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)..\left({x}+{n}\right)+… \\ $$$${to}\:{find}\:{a}_{\mathrm{0}\:} \:{put}\:{x}=\mathrm{0} \\ $$$${a}_{\mathrm{0}} ×{n}!=\mathrm{1}\:\:\:{a}_{\mathrm{0}} =\frac{\mathrm{1}}{{n}!} \\ $$$${similarly}\:{to}\:{find}\:{a}_{\mathrm{1}} \:{put}\:{x}+\mathrm{1}=\mathrm{0} \\ $$$${a}_{\mathrm{1}} \left(−\mathrm{1}\right)\left(\mathrm{1}\right)\left(\mathrm{2}\right)…\left({n}−\mathrm{1}\right)=\mathrm{1} \\ $$$${a}_{\mathrm{1}} ×\left(−\mathrm{1}\right)×\left({n}−\mathrm{1}\right)!=\mathrm{1} \\ $$$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\left(−\mathrm{1}\right)\left({n}−\mathrm{1}\right)!} \\ $$$${in}\:{this}\:{way}\:{w}\:{e}\:{find}\:{a}_{\mathrm{0}} ,{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} …{a}_{{n}} \\ $$$${so}\:{the}\:{intregation}\:{value}\:{is} \\ $$$${a}_{\mathrm{0}} {lnx}+{a}_{\mathrm{1}} {ln}\left({x}+\mathrm{1}\right)+{a}_{\mathrm{2}} \left({x}+\mathrm{2}\right)….+{a}_{{n}} \left({x}+{n}\right)+{c} \\ $$$${nxt}\:{step}\:{put}\:{value}\:{of}\:{a}_{\mathrm{0}} ,{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ….{a}_{{n}} \\ $$$${i}\:{have}\:{solved}\:{in}\:{detail}\:{for}\:{understanding}… \\ $$