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Question Number 129672 by Dwaipayan Shikari last updated on 17/Jan/21

$$\int \frac{dx}{{(1+x^{97})}^7}$$

Answered by mindispower last updated on 17/Jan/21

$$u=x^{97}\Rightarrow dx=\frac{u^{-\frac{96}{97}}}{97}du$$$$=\int \frac{u^{-\frac{96}{97}}}{97{(1+u)}^7}du...E$$$$\int \frac{x^s}{1+x}dx=\int \frac{x^s-x^{s+1}}{1-x^2}dx...F$$$$x^2=t$$$$=\frac{1}{2}\int \frac{1}{1-t}(t^{2s-\frac{1}{2}}-t^{2s+\frac{3}{2}})dt$$$$=\frac{1}{2}\int (\frac{1-t^{2s+\frac{3}{2}}}{1-t}-\frac{1-t^{2s-\frac{1}{2}}}{1-t})$$$$=\frac{1}{2}(\mathrm{\Psi }(2s+\frac{5}{2})-\mathrm{\Psi }(2s+\frac{1}{2}))$$$$iintegratebypart6time..E\mathrm{\&}F$$$$giveusanswr$$

Commented by Dwaipayan Shikari last updated on 17/Jan/21

$$Greatsir!CanitbefransformedintoHypergeometric$$$$functions?$$

Commented by mindispower last updated on 18/Jan/21

$$yes$$$$\frac{d^6}{{du}^6}.\frac{1}{(1+u)}=\frac{6!}{{(1+u)}^7}=\frac{d^6}{{du}^6}\sum _{k⩾0}{(-1)}^k{u}^k$$$$=\sum _{k⩾6}{(-1)}^{k}k(k-1)...(k-6).u^{k-6}$$$$=\sum _{k⩾0}{(-1)}^k.(k+6)......k.u^k$$$$\int \frac{dx}{{(1+x^{97})}^7}=\int \sum _{k⩾0}{(-x^{97})}^k.\frac{(k+6)...k}{1}dx,\mathrm{\forall }\mid x\mid <1$$$$=\sum _{k⩾0}\frac{{(-1)}^k{x}^{97k+1}.k...(k+6)}{(97k+1)}+c$$$$=x.(1+\sum _{k⩾1}\frac{{(-x^{97})}^k.(k+6)!}{(k-1)!.(k+\frac{1}{97)})}$$$$=x(1+\sum _{k⩾0}\frac{{(-x)}^{97k+97}.(k+7)!}{k!.(k+\frac{98}{97})})$$$$=x(1+x^{97}.\sum _{k⩾0}\frac{{(-x^{97})}^k}{k!}.\frac{{\left(8\right)}_k.{\left(\frac{1}{97}\right)}_k}{{\left(\frac{98}{97}\right)}_k})$$$$=x+x^{98}{.}_2{F}_1(8,\frac{1}{97};\frac{98}{97};-x^{97})+c$$$$$$$$$$

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