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Question Number 105565 by mathmax by abdo last updated on 30/Jul/20

let f (x) = x^{2} \ln (1 - x^{3})

1) calculate f^{(n)} (x) and f^{(n)} (0)

2) developp f at integr serie

3) calculate \int f (x) dx

Answered by mathmax by abdo last updated on 03/Aug/20

1) f (x) = x^{2} \ln (1 - x^{3}) \Rightarrow f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(x^{2})}^{(k)} {(\ln (1 - x^{3}))}^{(n - k)}

= C_{n}^{0} x^{2} {\ln (1 - x^{3})}^{(n)} + C_{n}^{1} (2 x) {\ln (1 - x^{3})}^{(n - 1)} + C_{n}^{2} 2 {\ln (1 - x^{3})}^{(n - 2)}

let find {\ln (1 - x^{3})}^{(m)} we have

{\ln (1 - x^{3})}^{(1)} = \frac{- {3 x}^{2}}{1 - x^{3}} = \frac{{3 x}^{2}}{x^{3} - 1} = \frac{{3 x}^{2}}{(x - 1) (x^{2} + x + 1)}

= \frac{{3 x}^{2}}{(x - 1) (x - e^{i \frac{2 π}{3}}) (x - e^{- \frac{i 2 π}{3}})} = \frac{a}{x - 1} + \frac{b}{x - e^{\frac{i 2 π}{3}}} + \frac{c}{(x - e^{- \frac{i 2 π}{3}})}

a = 1, b = \frac{{3 e}^{i \frac{4 π}{3}}}{(e^{\frac{i 2 π}{3}} - 1) (2 isin (\frac{2 π}{3}))} = \frac{{3 e}^{\frac{i 4 π}{3}}}{i \sqrt{3} (e^{\frac{i 2 π}{3}} - 1)}

b = \frac{{3 e}^{- \frac{i 4 π}{3}}}{(e^{- \frac{i 2 π}{3}} - 1) (- 2 i \sin (\frac{2 π}{3}))} = \frac{- 3 e^{- \frac{i 4 π}{3}}}{i \sqrt{3} (e^{- \frac{i 2 π}{3}} - 1)} so the coefficient are

known \Rightarrow {\ln (1 - x^{3})}^{(m)} = {(\frac{a}{x - 1})}^{(m - 1)} + {(\frac{b}{x - e^{\frac{i 2 π}{3}}})}^{(m - 1)}

+ {(\frac{c}{(x - e^{- \frac{i 2 π}{3}})})}^{(m - 1)} = a \frac{{(- 1)}^{m - 1} (m - 1)!}{{(x - 1)}^{m}} + \frac{b {(- 1)}^{m - 1} (m - 1)!}{(x - e^{\frac{i 2 π}{3}})}

+ \frac{c {(- 1)}^{m - 1} (m - 1)!}{(x - e^{- \frac{i 2 π}{3}})} = A_{n}

f^{(n)} (x) = x^{2} A_{n} (x) + 2 nx A_{n - 1} (x) + {2 C}_{n}^{2} A_{n - 2} (x)