Question-134052

Question Number 134052 by shaker last updated on 27/Feb/21

Answered by mathmax by abdo last updated on 27/Feb/21

I = \int \frac{x^{3}}{x^{6} + 3} dx \Rightarrow I = \int \frac{x^{3}}{x^{6} + {(3^{\frac{1}{6}})}^{6}} dx =_{x = 3^{\frac{1}{6}} t} \int \frac{3^{\frac{1}{2}} t^{3}}{3 (1 + t^{6})} 3^{\frac{1}{6}} dt

= 3^{\frac{1}{2} + \frac{1}{6} - 1} \int \frac{t^{3}}{t^{6} + 1} dt = 3^{- \frac{1}{3}} \int \frac{t^{3}}{t^{6} + 1} dt

z^{6} + 1 = 0 \Rightarrow z^{6} = e^{i (π + 2 k π)} = e^{i (2 k + 1) π} \Rightarrow z_{k} = e^{\frac{i (2 k + 1) π}{6}} and

k \in [[0, 5]] \Rightarrow \frac{t^{3}}{t^{6} + 1} = \frac{t^{3}}{\prod_{k = 0}^{5} (t - z_{k})} = \sum_{k = 0}^{5} \frac{a_{k}}{t - z_{k}}

a_{k} = \frac{z_{k}^{3}}{{6 z}_{k}^{5}} = \frac{z_{k}^{4}}{- 6} = - \frac{1}{6} z_{k}^{4} \Rightarrow \frac{t^{3}}{t^{6} + 1} = - \frac{1}{6} \sum_{k = 0}^{5} \frac{e^{\frac{i (2 k + 1) 2 π}{3}}}{t - e^{\frac{i (2 k + 1) π}{6}}} \Rightarrow

\int \frac{t^{3}}{1 + t^{6}} dt = - \frac{1}{6} \sum_{k = 0}^{5} e^{\frac{2 π i (2 k + 1)}{3}} \ln (t - e^{\frac{i (2 k + 1) π}{6}}) + C