Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 82283 by M±th+et£s last updated on 19/Feb/20

$$\int x^3\sqrt{x^3+1}dx$$

Commented by MJS last updated on 20/Feb/20

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{impossible}\mathrm{for}\mathrm{me}\mathrm{to}\mathrm{solve}\mathrm{this}$$

Commented by M±th+et£s last updated on 20/Feb/20

$$icantsolveittosir$$$$iposteditaskingforhelp$$$$butithinkthatitsspecialintegral$$

Answered by mind is power last updated on 20/Feb/20

$$Hello$$$$\sqrt{1+r}=1+\frac{r}{2}+\sum _{k⩾2}{(-1)}^{k-1}.\frac{(2k-3)!!}{2^{k}k!}.x^k$$$$=1+\sum _{k⩾1}{(-1)}^{k-1}\frac{(2k-2)!}{2^{2k-1}.k{((k-1)!)}^2}{x}^k$$$$\int x^3\sqrt{1+x^3}dx$$$$=\int x^3(1+\sum _{k⩾1}{(-1)}^{k-1}\frac{(2k-2)!}{2^{2k-1}k!.(k-1)!}.x^{3k})dx$$$$=\frac{x^4}{4}+\sum _{k⩾1}\frac{{(-1)}^{k-1}(2k-2)!4}{2^{2k-1}(k-1)!.(3k+4).k!}{x}^{3k+4}$$$$=\frac{x^4}{4}(1+\sum _{k⩾1}\frac{{(-1)}^{k-1}(2k-2)!}{2^{2k-1}(k-1)!(3k+4)}.\frac{x^{3k}}{k!})$$$$=\frac{x^4}{4}(1+\sum _{k⩾1}\frac{{(-1)}^{k-1}.4.(2k-3)!!}{2^k(k+4)}.\frac{x^{3k}}{k!})$$$$=\frac{x^4}{4}(1+\sum _{k⩾1}\frac{(-\frac{1}{2}).......(-\frac{1}{2}+k-1)\frac{4}{3}.....(k+\frac{4}{3}-1)}{\left(\frac{7}{3}\right)......(\frac{7}{3}+k-1)}.{\left(\frac{x^3}{2}\right)}^k.\frac{1}{k!}$$$$=\frac{x^4}{4}{}_2{F}_1(-\frac{1}{2},\frac{4}{3};\frac{7}{3};\left(\frac{x^3}{2}\right))+c$$

Commented by MJS last updated on 20/Feb/20

$$\mathrm{I}\mathrm{thought}\mathrm{there}\mathrm{was}\mathrm{a}\mathrm{possibility}\mathrm{using}\mathrm{series}$$$$\mathrm{but}\mathrm{I}\mathrm{have}\mathrm{never}\mathrm{learned}\mathrm{these}\mathrm{things}...$$

Commented by M±th+et£s last updated on 20/Feb/20

$$greatsirthankyousomuch.godblessyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com