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 ∫ f(x)dx


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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
	$1) f(x)=x^2 ln(1−x^3 ) ⇒f^((n)) (x) =Σ_(k=0) ^n C_n ^k (x^2 )^((k)) (ln(1−x^3 ))^((n−k)) =C_n ^0 x^2 {ln(1−x^3 )}^((n)) +C_n ^1 (2x){ln(1−x^3 )}^((n−1)) +C_n ^2 2{ln(1−x^3 )}^((n−2)) let find {ln(1−x^3 )}^((m)) we have {ln(1−x^3 )}^((1) ) =((−3x^2 )/(1−x^3 )) =((3x^2 )/(x^3 −1)) =((3x^2 )/((x−1)(x^2 +x+1))) =((3x^2 )/((x−1)(x−e^(i((2π)/3)) )(x−e^(−((i2π)/3)) ))) =(a/(x−1)) +(b/(x−e^((i2π)/3) )) +(c/((x−e^(−((i2π)/3)) ))) a =1 ,b =((3e^(i((4π)/3)) )/((e^((i2π)/3) −1)(2isin(((2π)/3))))) =((3e^((i4π)/3) )/(i(√3)(e^((i2π)/3) −1))) b =((3e^(−((i4π)/3)) )/((e^(−((i2π)/3)) −1)(−2i sin(((2π)/3))))) =((−3 e^(−((i4π)/3)) )/(i(√3)(e^(−((i2π)/3)) −1))) so the coefficient are known ⇒{ln(1−x^3 )}^((m)) =((a/(x−1)))^((m−1)) +((b/(x−e^((i2π)/3) )))^((m−1)) +((c/((x−e^(−((i2π)/3)) ))))^((m−1)) =a (((−1)^(m−1) (m−1)!)/((x−1)^m )) +((b(−1)^(m−1) (m−1)!)/((x−e^((i2π)/3) ))) +((c(−1)^(m−1) (m−1)!)/((x−e^(−((i2π)/3)) ))) =A_n f^((n)) (x) =x^2 A_n (x) +2nx A_(n−1) (x) +2C_n ^2 A_(n−2) (x)$
	$1) f (x) = x^{2} \ln (1 - x^{3}) \Rightarrow f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(x^{2})}^{(k)} {(\ln (1 - x^{3}))}^{(n - k)}$ $= C_{n}^{0} x^{2} {\ln (1 - x^{3})}^{(n)} + C_{n}^{1} (2 x) {\ln (1 - x^{3})}^{(n - 1)} + C_{n}^{2} 2 {\ln (1 - x^{3})}^{(n - 2)}$ $let find {\ln (1 - x^{3})}^{(m)} we have$ ${\ln (1 - x^{3})}^{(1)} = \frac{- {3 x}^{2}}{1 - x^{3}} = \frac{{3 x}^{2}}{x^{3} - 1} = \frac{{3 x}^{2}}{(x - 1) (x^{2} + x + 1)}$ $= \frac{{3 x}^{2}}{(x - 1) (x - e^{i \frac{2 π}{3}}) (x - e^{- \frac{i 2 π}{3}})} = \frac{a}{x - 1} + \frac{b}{x - e^{\frac{i 2 π}{3}}} + \frac{c}{(x - e^{- \frac{i 2 π}{3}})}$ $a = 1, b = \frac{{3 e}^{i \frac{4 π}{3}}}{(e^{\frac{i 2 π}{3}} - 1) (2 isin (\frac{2 π}{3}))} = \frac{{3 e}^{\frac{i 4 π}{3}}}{i \sqrt{3} (e^{\frac{i 2 π}{3}} - 1)}$ $b = \frac{{3 e}^{- \frac{i 4 π}{3}}}{(e^{- \frac{i 2 π}{3}} - 1) (- 2 i \sin (\frac{2 π}{3}))} = \frac{- 3 e^{- \frac{i 4 π}{3}}}{i \sqrt{3} (e^{- \frac{i 2 π}{3}} - 1)} so the coefficient are$ $known \Rightarrow {\ln (1 - x^{3})}^{(m)} = {(\frac{a}{x - 1})}^{(m - 1)} + {(\frac{b}{x - e^{\frac{i 2 π}{3}}})}^{(m - 1)}$ $+ {(\frac{c}{(x - e^{- \frac{i 2 π}{3}})})}^{(m - 1)} = a \frac{{(- 1)}^{m - 1} (m - 1)!}{{(x - 1)}^{m}} + \frac{b {(- 1)}^{m - 1} (m - 1)!}{(x - e^{\frac{i 2 π}{3}})}$ $+ \frac{c {(- 1)}^{m - 1} (m - 1)!}{(x - e^{- \frac{i 2 π}{3}})} = A_{n}$ $f^{(n)} (x) = x^{2} A_{n} (x) + 2 nx A_{n - 1} (x) + {2 C}_{n}^{2} A_{n - 2} (x)$
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