$$\int_{\2}^{\9}(\ $\sqrt{\8+x^3}$ ) d\x $\infty$ $$

asked by guest
on Nov 18, 2024 at 6:49 am



You asked:

Evaluate the expression: ∫298+x3dx⋅∞\int_{2}^{9}{\sqrt{8 + {x}^{3}}}dx \cdot \infty

MathBot Answer:

Evaluated



∫298+x3dx⋅∞=∞(−42Γ(13)2F1(−12,1343|−1)3Γ(43)+62Γ(13)2F1(−12,1343|729eiπ8)Γ(43))\displaystyle \int_{2}^{9}{\sqrt{8 + {x}^{3}}}dx \cdot \infty = \infty \left(- \frac{4 \sqrt{2} \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {-1} \right)}}{3 \Gamma\left(\frac{4}{3}\right)} + \frac{6 \sqrt{2} \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {\frac{729 e^{i \pi}}{8}} \right)}}{\Gamma\left(\frac{4}{3}\right)}\right)

Expanded

∫298+x3dx⋅∞=∞∫29x3+8 dx\int_{2}^{9}{\sqrt{8 + {x}^{3}}}dx \cdot \infty = \infty \int\limits_{2}^{9} \sqrt{x^{3} + 8}\, dx

Factored

∫298+x3dx⋅∞=∞∫29x3+8 dx\int_{2}^{9}{\sqrt{8 + {x}^{3}}}dx \cdot \infty = \infty \int\limits_{2}^{9} \sqrt{x^{3} + 8}\, dx

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