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Question Number 86454 by redmiiuser last updated on 28/Mar/20

$$plscheckthe$$$$questionbelow$$

Answered by redmiiuser last updated on 28/Mar/20

$$\int 1/\sqrt{}(1+x^3).dx$$$${(1+x^3)}^{(-1/2)}$$$$=1+(-1/2)x^3+(-1/2)(-3/2)x^6/2!+(-1/2)(-3/2)(-5/2)x^9/3!+(-1/2)(-3/2)(-5/2)(-7/2)/4!x^{12}+...$$$$$$$$therefore$$$${(1+x^3)}^{(-1/2)}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(({(-1)}^n.\left(2n\right)!.{\left(x^3\right)}^n)/\left((2^{2n}.{(n!)}^2)\right)\right)$$$$\therefore \int {(1+x^3)}^{(-1/2)}.dx$$$$=\int \underset{n=0}{\sum ^{\mathrm{\infty }}}\left(({(-1)}^n.\left(2n\right)!.{\left(x^3\right)}^n)/(2^{2n}.{(n!)}^2)\right).dx$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(({(-1)}^n.\left(2n\right)!.{\left(x\right)}^{3n+1})/((3n+1).\left(2^{2n}\right).{(n!)}^2)\right)$$$$$$

Commented by redmiiuser last updated on 29/Mar/20

$$cananyonecomment$$$$whethertheabove$$$$processiscorrector$$$$wrong.$$

Commented by redmiiuser last updated on 29/Mar/20

$$mr.Tanmaycanyou$$$$helpmeinthe$$$$aboveproblempls.$$

Commented by TANMAY PANACEA. last updated on 29/Mar/20

$$\int \frac{dx}{\sqrt{1+x^3}}$$$$x^3={tan}^2\alpha$$$$x={\left(tan\alpha \right)}^{\frac{2}{3}}\to dx=\frac{2}{3}\times {\left(tan\alpha \right)}^{\frac{2}{3}-1}\times {sec}^2\alpha d\alpha$$$$\int \frac{2{\left(tan\alpha \right)}^{\frac{-1}{3}}\times {sec}^2\alpha \times d\alpha }{3\times {(1+{tan}^2\alpha )}^{\frac{1}{2}}}$$$$\frac{2}{3}\int \frac{sec\alpha d\alpha }{{\left(tan\alpha \right)}^{\frac{1}{3}}}$$$$\frac{2}{3}\int \frac{d\alpha }{{\left(sin\alpha \right)}^{\frac{1}{3}}{\left(cos\alpha \right)}^{\frac{2}{3}}}$$

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