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Question Number 107515 by mohammad17 last updated on 11/Aug/20
∫(x^4 /(1+x^8 ))dx
$$\int\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{8}} }{dx} \\ $$
Answered by Ar Brandon last updated on 11/Aug/20
I=∫(x^4 /(1+x^8 ))dx Let x^8 +1=0 ⇒ x_k ^8 =e^((2k+1)iπ) ⇒ x_k =e^((((2k+1)/8))iπ) k∈[0,7] ⇒x^8 +1=Π_(k=0) ^7 (x−x_k ) ⇒∫(x^4 /(1+x^8 ))dx=∫(x^4 /(Π_(k=0) ^7 (x−x_k )))dx=∫{Σ_(k=0) ^7 (a_k /(x−x_k ))dx} a_k =(x_k ^4 /(8x_k ^7 ))=−(x_k ^5 /8) ⇒∫(x^4 /(1+x^8 ))dx=−(1/8)Σ_(k=0) ^7 {∫(x_k ^5 /(x−x_k ))dx} ⇒∫(x^4 /(1+x^8 ))dx=−(1/8)[Σ_(k=0) ^7 x_k ^5 ln∣x−x_k ∣]+C, x_k =e^((((2k+1)/8))iπ)
$$\mathrm{I}=\int\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{8}} }\mathrm{dx} \\ $$$$\mathrm{Let}\:\mathrm{x}^{\mathrm{8}} +\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{x}_{\mathrm{k}} ^{\mathrm{8}} =\mathrm{e}^{\left(\mathrm{2k}+\mathrm{1}\right)\mathrm{i}\pi} \:\Rightarrow\:\mathrm{x}_{\mathrm{k}} =\mathrm{e}^{\left(\frac{\mathrm{2k}+\mathrm{1}}{\mathrm{8}}\right)\mathrm{i}\pi} \mathrm{k}\in\left[\mathrm{0},\mathrm{7}\right] \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{8}} +\mathrm{1}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{7}} {\prod}}\left(\mathrm{x}−\mathrm{x}_{\mathrm{k}} \right) \\ $$$$\Rightarrow\int\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{8}} }\mathrm{dx}=\int\frac{\mathrm{x}^{\mathrm{4}} }{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{7}} {\prod}}\left(\mathrm{x}−\mathrm{x}_{\mathrm{k}} \right)}\mathrm{dx}=\int\left\{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{7}} {\sum}}\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{x}−\mathrm{x}_{\mathrm{k}} }\mathrm{dx}\right\} \\ $$$$\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{4}} }{\mathrm{8x}_{\mathrm{k}} ^{\mathrm{7}} }=−\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{5}} }{\mathrm{8}} \\ $$$$\Rightarrow\int\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{8}} }\mathrm{dx}=−\frac{\mathrm{1}}{\mathrm{8}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{7}} {\sum}}\left\{\int\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{5}} }{\mathrm{x}−\mathrm{x}_{\mathrm{k}} }\mathrm{dx}\right\} \\ $$$$\Rightarrow\int\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{8}} }\mathrm{dx}=−\frac{\mathrm{1}}{\mathrm{8}}\left[\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{7}} {\sum}}\mathrm{x}_{\mathrm{k}} ^{\mathrm{5}} \mathrm{ln}\mid\mathrm{x}−\mathrm{x}_{\mathrm{k}} \mid\right]+\mathcal{C},\:\:\mathrm{x}_{\mathrm{k}} =\mathrm{e}^{\left(\frac{\mathrm{2k}+\mathrm{1}}{\mathrm{8}}\right)\mathrm{i}\pi} \\ $$
Commented by mohammad17 last updated on 11/Aug/20
thank you sir . can you solve by different method ?
$${thank}\:{you}\:{sir}\:.\:{can}\:{you}\:{solve}\:{by}\:{different}\:{method}\:? \\ $$
Commented by Ar Brandon last updated on 11/Aug/20
Not yet. Do you face difficulties with this ? ��
Commented by Ar Brandon last updated on 11/Aug/20
Methode Complexe pour resoudre les integrales Soit F(x)=((P(x))/(Q(x))) , degP(x)<degQ(x) et Q(x)∈C ie Q(x)=λΠ_(k=i) ^n (x−x_k ) alors F(x)=Σ_(k=i) ^n (a_k /(x−x_k )) d′ou^� a_k =((P(x))/(Q′(x))) Exemple I=∫(x^6 /(1+x^(12) ))dx x^(12) +1=0⇒x^(12) =−1=e^((2k+1)iπ) ⇒x_k =e^(((2k+1)/(12))iπ) ⇒I=∫(x^6 /(Π_(k=0) ^(11) (x−x_k )))dx=∫Σ_(k=0) ^(11) (a_k /(x−x_k ))dx a_k =(x_k ^6 /(12x_k ^(11) ))=(x_k ^7 /(12x_k ^(12) ))=(x_k ^7 /(−12)) {x_k ^(12) =e^((2k+1)iπ) =−1} ⇒I=((−1)/(12))∫Σ_(k=0) ^(11) (x_k ^7 /(x−x_k ))dx=((−1)/(12))Σ_(k=0) ^(11) x_k ^7 ln∣x−x_k ∣+C
$$\boldsymbol{\mathrm{Methode}}\:\boldsymbol{\mathrm{Complexe}}\:\boldsymbol{\mathrm{pour}}\:\boldsymbol{\mathrm{resoudre}}\:\boldsymbol{\mathrm{les}}\:\boldsymbol{\mathrm{integrales}} \\ $$$$\mathrm{Soit}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{P}\left(\mathrm{x}\right)}{\mathrm{Q}\left(\mathrm{x}\right)}\:,\:\mathrm{degP}\left(\mathrm{x}\right)<\mathrm{degQ}\left(\mathrm{x}\right)\:\mathrm{et}\:\mathrm{Q}\left(\mathrm{x}\right)\in\mathbb{C} \\ $$$$\mathrm{ie}\:\mathrm{Q}\left(\mathrm{x}\right)=\lambda\underset{\mathrm{k}=\mathrm{i}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{x}−\mathrm{x}_{\mathrm{k}} \right)\:\mathrm{alors} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\underset{\mathrm{k}=\mathrm{i}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{x}−\mathrm{x}_{\mathrm{k}} }\:\:\mathrm{d}'\mathrm{o}\grave {\mathrm{u}}\:\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{P}\left(\mathrm{x}\right)}{\mathrm{Q}'\left(\mathrm{x}\right)} \\ $$$$\mathrm{Exemple} \\ $$$$\mathcal{I}=\int\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{1}+\mathrm{x}^{\mathrm{12}} }\mathrm{dx}\:\: \\ $$$$\mathrm{x}^{\mathrm{12}} +\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{12}} =−\mathrm{1}=\mathrm{e}^{\left(\mathrm{2k}+\mathrm{1}\right)\mathrm{i}\pi} \Rightarrow\mathrm{x}_{\mathrm{k}} =\mathrm{e}^{\frac{\mathrm{2k}+\mathrm{1}}{\mathrm{12}}\mathrm{i}\pi} \\ $$$$\Rightarrow\mathcal{I}=\int\frac{\mathrm{x}^{\mathrm{6}} }{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{11}} {\prod}}\left(\mathrm{x}−\mathrm{x}_{\mathrm{k}} \right)}\mathrm{dx}=\int\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{11}} {\sum}}\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{x}−\mathrm{x}_{\mathrm{k}} }\mathrm{dx} \\ $$$$\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{6}} }{\mathrm{12x}_{\mathrm{k}} ^{\mathrm{11}} }=\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{7}} }{\mathrm{12x}_{\mathrm{k}} ^{\mathrm{12}} }=\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{7}} }{−\mathrm{12}}\:\left\{\mathrm{x}_{\mathrm{k}} ^{\mathrm{12}} =\mathrm{e}^{\left(\mathrm{2k}+\mathrm{1}\right)\mathrm{i}\pi} =−\mathrm{1}\right\} \\ $$$$\Rightarrow\mathcal{I}=\frac{−\mathrm{1}}{\mathrm{12}}\int\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{11}} {\sum}}\frac{\mathrm{x}_{\mathrm{k}} ^{\mathrm{7}} }{\mathrm{x}−\mathrm{x}_{\mathrm{k}} }\mathrm{dx}=\frac{−\mathrm{1}}{\mathrm{12}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{11}} {\sum}}\mathrm{x}_{\mathrm{k}} ^{\mathrm{7}} \mathrm{ln}\mid\mathrm{x}−\mathrm{x}_{\mathrm{k}} \mid+\mathcal{C} \\ $$
Commented by mohammad17 last updated on 11/Aug/20
thank you sir very thank
$${thank}\:{you}\:{sir}\:{very}\:{thank} \\ $$
Answered by Sarah85 last updated on 11/Aug/20
x^8 +1= (x^2 −(√(2−(√2)))x+1)(x^2 +(√(2−(√2)))x+1)× ×(x^2 −(√(2+(√2)))x+1)(x^2 +(√(2+(√2)))x+1) now decompose, which is no fun to write out let α=(√(2−(√2)))∧β=(√(2+(√2))) (x^4 /((x^2 −αx+1)(x^2 +αx+1)(x^2 −βx+1)(x^2 +βx+1)))= =((ax+b)/(x^2 −αx+1))+((cx+d)/(x^2 +αx+1))+((ex+f)/(x^2 −βx+1))+((gx+h)/(x^2 +βx+1)) I don′t want to do this job for you
$${x}^{\mathrm{8}} +\mathrm{1}= \\ $$$$\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}{x}+\mathrm{1}\right)× \\ $$$$×\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}{x}+\mathrm{1}\right) \\ $$$$\mathrm{now}\:\mathrm{decompose},\:\mathrm{which}\:\mathrm{is}\:\mathrm{no}\:\mathrm{fun}\:\mathrm{to}\:\mathrm{write}\:\mathrm{out} \\ $$$$\mathrm{let}\:\alpha=\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\wedge\beta=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$$$\frac{{x}^{\mathrm{4}} }{\left({x}^{\mathrm{2}} −\alpha{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\alpha{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\beta{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\beta{x}+\mathrm{1}\right)}= \\ $$$$=\frac{{ax}+{b}}{{x}^{\mathrm{2}} −\alpha{x}+\mathrm{1}}+\frac{{cx}+{d}}{{x}^{\mathrm{2}} +\alpha{x}+\mathrm{1}}+\frac{{ex}+{f}}{{x}^{\mathrm{2}} −\beta{x}+\mathrm{1}}+\frac{{gx}+{h}}{{x}^{\mathrm{2}} +\beta{x}+\mathrm{1}} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{want}\:\mathrm{to}\:\mathrm{do}\:\mathrm{this}\:\mathrm{job}\:\mathrm{for}\:\mathrm{you} \\ $$

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