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

Question Number 145378 by mnjuly1970 last updated on 04/Jul/21

\int_{0}^{\infty} (\sqrt{1 + x^{4}} - x^{2}) dx = ?

Answered by qaz last updated on 05/Jul/21

\int_{0}^{\infty} (\sqrt{1 + x^{4}} - x^{2}) dx

= x (\sqrt{1 + x^{4}} - x^{2}) ∣_{0}^{\infty} - 2 \int_{0}^{\infty} (\frac{1 + x^{4} - 1}{\sqrt{1 + x^{4}}} - x^{2}) dx

= \frac{x}{\sqrt{1 + x^{4}} + x^{2}} ∣_{0}^{\infty} - 2 \int_{0}^{\infty} (\sqrt{1 + x^{4}} - x^{2}) dx + 2 \int_{0}^{\infty} \frac{dx}{\sqrt{1 + x^{4}}}

= \frac{2}{3} \int_{0}^{\infty} \frac{dx}{\sqrt{1 + x^{4}}}

= \frac{2}{3} \cdot \frac{1}{4} \int_{0}^{\infty} \frac{u^{- 3 / 4}}{{(1 + u)}^{1 / 2}} du \dots \dots \dots \dots x^{4} \to u

= \frac{1}{6} B (\frac{1}{4}, \frac{1}{4})

= \frac{Γ^{2} (\frac{1}{4})}{6 \sqrt{π}}

Commented by mnjuly1970 last updated on 05/Jul/21

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