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Question Number 57229 by maxmathsup by imad last updated on 31/Mar/19
give ∫_0 ^1 (x^5 /(x^3 +1)) dx at form of serie
$${give}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{5}} }{{x}^{\mathrm{3}} \:+\mathrm{1}}\:{dx}\:{at}\:{form}\:{of}\:{serie} \\ $$
Commented by maxmathsup by imad last updated on 01/Apr/19
let I =∫_0 ^1 (x^5 /(x^3 +1)) dx ⇒I =∫_0 ^1 x^5 (Σ_(n=0) ^∞ (−x)^(3n) )dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(3n+5) dx =Σ_(n=0) ^∞ (−1)^n (1/(3n+6)) =(1/3) Σ_(n=0) ^∞ (((−1)^n )/(n+2)) let find the value of I we have Σ_(n=0) ^∞ (((−1)^n )/(n+2)) =_(n+2=p) Σ_(p=2) ^∞ (((−1)^(p−2) )/p) =Σ_(p=2) ^∞ (((−1)^p )/p) =Σ_(p=1) ^∞ (((−1)^p )/p) +1 =−ln(2) +1 ⇒I=(1/3)(1−ln(2))
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{5}} }{{x}^{\mathrm{3}} \:+\mathrm{1}}\:{dx}\:\:\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{5}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{x}\right)^{\mathrm{3}{n}} \right){dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{3}{n}+\mathrm{5}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{6}}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}} \\ $$$${let}\:{find}\:{the}\:{value}\:{of}\:{I}\:\:{we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{2}}\:=_{{n}+\mathrm{2}={p}} \:\sum_{{p}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{2}} }{{p}} \\ $$$$=\sum_{{p}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{{p}}\:=\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{{p}}\:+\mathrm{1}\:=−{ln}\left(\mathrm{2}\right)\:+\mathrm{1}\:\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−{ln}\left(\mathrm{2}\right)\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19
t=1+x^3 dt=3x^2 dx→(dt/3)=x^2 dx ∫_0 ^1 ((x^3 ×x^2 dx)/(1+x^3 )) ∫_1 ^2 ((t−1)/t)×(dt/3) (1/3)∫_1 ^2 (1−(1/t))dt (1/3)∣t−lnt∣_1 ^2 (1/3){(2−1)−(ln2−ln1)} (1/3){1−ln2)
$${t}=\mathrm{1}+{x}^{\mathrm{3}} \\ $$$${dt}=\mathrm{3}{x}^{\mathrm{2}} {dx}\rightarrow\frac{{dt}}{\mathrm{3}}={x}^{\mathrm{2}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} ×{x}^{\mathrm{2}} {dx}}{\mathrm{1}+{x}^{\mathrm{3}} } \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{t}−\mathrm{1}}{{t}}×\frac{{dt}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{t}}\right){dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mid{t}−{lnt}\mid_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left\{\left(\mathrm{2}−\mathrm{1}\right)−\left({ln}\mathrm{2}−{ln}\mathrm{1}\right)\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left\{\mathrm{1}−{ln}\mathrm{2}\right) \\ $$

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