1)find f(x)=∫_0 ^1 ln(1+xt^3 )dt with ∣x∣<1 2) calculate ∫


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Question Number 45635 by maxmathsup by imad last updated on 14/Oct/18

$1) f i n d f (x) = \int_{0}^{1} l n (1 + {x t}^{3}) d t w i t h ∣ x ∣< 1$ $2) c a l c u l a t e \int_{0}^{1} l n (2 + t^{3}) d t .$
Commented bymaxmathsup by imad last updated on 17/Oct/18
$1) changement^3 (√x)t =u give f(x)= ∫_0 ^3_(√x) ln(1+u^3 ) (du/((^3 (√x)))) =(1/((^3 (√x)))) ∫_0 ^3_(√x) ln(1+u^3 )du but we have proved that ∫ ln(1+x^3 )dx =xln(1+x^3 )−3x +ln∣x+1∣ −(1/2)ln(x^2 −x+1) +(√3)arctan(((2x−1)/(√3))) f(x)= (1/((^3 (√x))))[uln(1+u^3 )−3u +ln∣u+1∣−(1/2)ln(u^2 −u+1)+(√3)arctan(((2x−1)/(√3)))]_0 ^3_(√x) =(1/((^3 (√x)))){^3 (√x)ln(1+x)−3^ (^3 (√x))+ln∣1+^3 (√x)∣−(1/2)ln{ u^(2/3) −^3 (√x)+1} +(√3)arctan(((2(^3 (√x))−1)/(√3))) +((π(√3))/6)} .$
$1) c h a n g e m e n t^{3} \sqrt{x} t = u g i v e$ $f (x) = \int_{0}^{3_{\sqrt{x}}} l n (1 + u^{3}) \frac{d u}{(^{3} \sqrt{x})} = \frac{1}{(^{3} \sqrt{x})} \int_{0}^{3_{\sqrt{x}}} l n (1 + u^{3}) d u b u t w e h a v e p r o v e d t h a t$ $\int l n (1 + x^{3}) d x = x l n (1 + x^{3}) - 3 x + l n ∣ x + 1 ∣ - \frac{1}{2} l n (x^{2} - x + 1) + \sqrt{3} a r c t a n (\frac{2 x - 1}{\sqrt{3}})$ $f (x) = \frac{1}{(^{3} \sqrt{x})} {[u l n (1 + u^{3}) - 3 u + l n ∣ u + 1 ∣ - \frac{1}{2} l n (u^{2} - u + 1) + \sqrt{3} a r c t a n (\frac{2 x - 1}{\sqrt{3}})]}_{0}^{3_{\sqrt{x}}}$ $= \frac{1}{(^{3} \sqrt{x})} {^{3} \sqrt{x} l n (1 + x) - 3^{} (^{3} \sqrt{x}) + l n ∣ 1 +^{3} \sqrt{x} ∣ - \frac{1}{2} l n {u^{\frac{2}{3}} -^{3} \sqrt{x} + 1}$ $+ \sqrt{3} a r c t a n (\frac{2 (^{3} \sqrt{x}) - 1}{\sqrt{3}}) + \frac{π \sqrt{3}}{6}} .$
Commented bymaxmathsup by imad last updated on 17/Oct/18
$let A =∫_0 ^1 ln(2+t^3 )dt ⇒ A = ∫_0 ^1 ln{2(1 +(1/2)t^3 )}dt =ln(2) +∫_0 ^1 ln(1+(1/2)t^3 )dt =ln(2)+f((1/2)) =ln(2) +^3 (√2){^3 (√2)ln((3/2))−3(^3 (√2)) +ln∣1+^3 (√2)∣ −(1/2)ln{ 2^(−(2/3)) −(1/((^3 (√x)))) +1} +(√3)arctan((((2/((^3 (√2))))−1)/(√3)))+((π(√3))/6)} .$
$l e t A = \int_{0}^{1} l n (2 + t^{3}) d t \Rightarrow A = \int_{0}^{1} l n {2 (1 + \frac{1}{2} t^{3})} d t$ $= l n (2) + \int_{0}^{1} l n (1 + \frac{1}{2} t^{3}) d t$ $= l n (2) + f (\frac{1}{2}) = l n (2) +^{3} \sqrt{2} {^{3} \sqrt{2} l n (\frac{3}{2}) - 3 (^{3} \sqrt{2}) + l n ∣ 1 +^{3} \sqrt{2} ∣$ $- \frac{1}{2} l n {2^{- \frac{2}{3}} - \frac{1}{(^{3} \sqrt{x})} + 1} + \sqrt{3} a r c t a n (\frac{\frac{2}{(^{3} \sqrt{2})} - 1}{\sqrt{3}}) + \frac{π \sqrt{3}}{6}} .$
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