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Question Number 119894 by danielasebhofoh last updated on 27/Oct/20

Answered by mathmax by abdo last updated on 28/Oct/20

  let I_n =∫ (dx/(x(x^n −a^n )))  if a≠0 we do the changement x=at ⇒  I_n =∫ ((adt)/(at(a^n t^n −a^n ))) =(1/a^n )∫  (dt/(t(t^n −1))) let decompose  F(t)=(1/(t(t^n −1))) inside C(x)  t^n −1 =0 ⇒_ t_k =e^(i((2kπ)/n))   and k∈[[0,n−1]] ⇒  F(t) =(1/(tΠ_(k=0) ^(n−1) (t−t_k ))) =(c/t) +Σ_(k=0) ^(n−1)  (a_k /(t−t_k ))  c=−1 ⇒F(t) =−(1/t) +Σ_(k=0) ^(n−1)  (a_k /(t−t_k )) ⇒  F(t)+(1/t) =(1/(t(t^n −1)))+(1/t) =(1/t){(1/(t^n −1))+1} =(1/t)×(t^n /(t^n −1)) =(t^(n−1) /(t^n −1))   ⇒a_k =(t_k ^(n−1) /(n t_k ^(n−1) )) =(1/n) ⇒F(t) =−(1/t)+(1/n)Σ_(k=0) ^(n−1)  (1/(t−t_k )) ⇒  ∫ F(t)dt =−ln∣t∣ +(1/n)Σ_(k=0) ^(n−1) ln(t−e^((i2kπ)/n) ) +C ⇒  I_n =(1/a^n ){−ln∣(x/a)∣+(1/n)Σ_(k=0) ^(n−1) ln((x/a)−e^((i2kπ)/n) )} +C

$$\:\:\mathrm{let}\:\mathrm{I}_{\mathrm{n}} =\int\:\frac{\mathrm{dx}}{\mathrm{x}\left(\mathrm{x}^{\mathrm{n}} −\mathrm{a}^{\mathrm{n}} \right)}\:\:\mathrm{if}\:\mathrm{a}\neq\mathrm{0}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\mathrm{at}\:\Rightarrow \\ $$$$\mathrm{I}_{\mathrm{n}} =\int\:\frac{\mathrm{adt}}{\mathrm{at}\left(\mathrm{a}^{\mathrm{n}} \mathrm{t}^{\mathrm{n}} −\mathrm{a}^{\mathrm{n}} \right)}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{n}} }\int\:\:\frac{\mathrm{dt}}{\mathrm{t}\left(\mathrm{t}^{\mathrm{n}} −\mathrm{1}\right)}\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\mathrm{t}\left(\mathrm{t}^{\mathrm{n}} −\mathrm{1}\right)}\:\mathrm{inside}\:\mathrm{C}\left(\mathrm{x}\right) \\ $$$$\mathrm{t}^{\mathrm{n}} −\mathrm{1}\:=\mathrm{0}\:\Rightarrow_{} \mathrm{t}_{\mathrm{k}} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{2k}\pi}{\mathrm{n}}} \:\:\mathrm{and}\:\mathrm{k}\in\left[\left[\mathrm{0},\mathrm{n}−\mathrm{1}\right]\right]\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\mathrm{1}}{\mathrm{t}\prod_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \left(\mathrm{t}−\mathrm{t}_{\mathrm{k}} \right)}\:=\frac{\mathrm{c}}{\mathrm{t}}\:+\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{t}−\mathrm{t}_{\mathrm{k}} } \\ $$$$\mathrm{c}=−\mathrm{1}\:\Rightarrow\mathrm{F}\left(\mathrm{t}\right)\:=−\frac{\mathrm{1}}{\mathrm{t}}\:+\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{t}−\mathrm{t}_{\mathrm{k}} }\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{t}}\:=\frac{\mathrm{1}}{\mathrm{t}\left(\mathrm{t}^{\mathrm{n}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{t}}\:=\frac{\mathrm{1}}{\mathrm{t}}\left\{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{n}} −\mathrm{1}}+\mathrm{1}\right\}\:=\frac{\mathrm{1}}{\mathrm{t}}×\frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{t}^{\mathrm{n}} −\mathrm{1}}\:=\frac{\mathrm{t}^{\mathrm{n}−\mathrm{1}} }{\mathrm{t}^{\mathrm{n}} −\mathrm{1}} \\ $$$$\:\Rightarrow\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{t}_{\mathrm{k}} ^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}\:\mathrm{t}_{\mathrm{k}} ^{\mathrm{n}−\mathrm{1}} }\:=\frac{\mathrm{1}}{\mathrm{n}}\:\Rightarrow\mathrm{F}\left(\mathrm{t}\right)\:=−\frac{\mathrm{1}}{\mathrm{t}}+\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{t}−\mathrm{t}_{\mathrm{k}} }\:\Rightarrow \\ $$$$\int\:\mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}\:=−\mathrm{ln}\mid\mathrm{t}\mid\:+\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \mathrm{ln}\left(\mathrm{t}−\mathrm{e}^{\frac{\mathrm{i2k}\pi}{\mathrm{n}}} \right)\:+\mathrm{C}\:\Rightarrow \\ $$$$\mathrm{I}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{n}} }\left\{−\mathrm{ln}\mid\frac{\mathrm{x}}{\mathrm{a}}\mid+\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{a}}−\mathrm{e}^{\frac{\mathrm{i2k}\pi}{\mathrm{n}}} \right)\right\}\:+\mathrm{C} \\ $$

Answered by TANMAY PANACEA last updated on 28/Oct/20

simple problem  ∫((x^(n−1) dx)/(x^n (x^n −a^n ))) [multiply N_r and D_r by x^(n−1) ]  t=x^n →(dt/dx)=nx^(n−1)   ∫(dt/(n(t)(t−a^n )))  (1/(na^n ))∫((1/(t−a^n ))−(1/t))dt  (1/(na^n ))ln(((t−a^n )/t))+c→(1/(na^n ))ln(((x^n −a^n )/x^n ))+c

$${simple}\:{problem} \\ $$$$\int\frac{{x}^{{n}−\mathrm{1}} {dx}}{{x}^{{n}} \left({x}^{{n}} −{a}^{{n}} \right)}\:\left[\boldsymbol{{multiply}}\:\boldsymbol{{N}}_{{r}} {and}\:{D}_{{r}} {by}\:{x}^{{n}−\mathrm{1}} \right] \\ $$$${t}={x}^{{n}} \rightarrow\frac{{dt}}{{dx}}={nx}^{{n}−\mathrm{1}} \\ $$$$\int\frac{{dt}}{{n}\left({t}\right)\left({t}−{a}^{{n}} \right)} \\ $$$$\frac{\mathrm{1}}{{na}^{{n}} }\int\left(\frac{\mathrm{1}}{{t}−{a}^{{n}} }−\frac{\mathrm{1}}{{t}}\right){dt} \\ $$$$\frac{\mathrm{1}}{{na}^{{n}} }{ln}\left(\frac{{t}−{a}^{{n}} }{{t}}\right)+{c}\rightarrow\frac{\mathrm{1}}{\boldsymbol{{na}}^{\boldsymbol{{n}}} }\boldsymbol{{ln}}\left(\frac{\boldsymbol{{x}}^{\boldsymbol{{n}}} −\boldsymbol{{a}}^{\boldsymbol{{n}}} }{\boldsymbol{{x}}^{\boldsymbol{{n}}} }\right)+\boldsymbol{{c}} \\ $$

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