find-0-1-x-2-x-4-1-dx-

Question Number 130725 by mathmax by abdo last updated on 28/Jan/21

find \int_{0}^{1} \frac{x^{2}}{\sqrt{x^{4} + 1}} dx

Answered by Dwaipayan Shikari last updated on 28/Jan/21

\frac{1}{\sqrt{1 + x}} = \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n!} {(- 1)}^{n} x^{n}

\frac{x^{2}}{\sqrt{1 + x^{4}}} = \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n!} {(- 1)}^{n} x^{4 n + 2}

\int_{0}^{1} \frac{x^{2}}{\sqrt{1 + x^{4}}} = \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n! (4 n + 3)} {(- 1)}^{n} = \frac{1}{3} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n} {(\frac{3}{4})}_{n}}{{(\frac{7}{4})}_{n} n!} {(- 1)}^{n}

A s, \frac{1}{4 n + 3} = \frac{1}{4} (\frac{1}{n + \frac{3}{4}}) = \frac{1}{4} (\frac{Γ (n + \frac{3}{4})}{Γ (n + \frac{7}{4})}) = \frac{4}{3} . \frac{1}{4} . \frac{{(\frac{3}{4})}_{n}}{{(\frac{7}{4})}_{n}}

= \frac{1}{3}_{2} F_{1} (\frac{1}{2}, \frac{3}{4}; \frac{7}{4}; - 1)

Commented by mathmax by abdo last updated on 28/Jan/21

thank you sir .