Question Number 102470 by  M±th+et+s last updated on 09/Jul/20

∫(dx/(x^(10) +x^2 ))

Answered by Ar Brandon last updated on 09/Jul/20

I=∫(dx/(x^(10) +x^2 ))=∫(dx/(x^2 (x^8 +1)))  (1/(x^2 (x^8 +1)))=((ax+b)/x^2 )+((cx^7 +dx^6 +ex^5 +fx^4 +gx^3 +hx^2 +ix+j)/(x^8 +1))  x=0 ⇒b=1 , a+c=0 , d+b=0 ⇒d=−1  e=0 , f=0 , g=0 , h=0 , i=0 , j=0 , a=0 ⇒c=0  (1/(x^2 (x^8 +1)))=(1/x^2 )−(x^6 /(x^8 +1))  ⇒I=∫{(1/x^2 )−(x^6 /(x^8 +1))}dx=−(1/x)−∫(x^6 /(x^8 +1))dx  J=∫(x^6 /(x^8 +1))dx , x^8 +1=0 ⇒x^8 =e^((2k+1)πi) ⇒x_k =e^((((2k+1)πi)/8) )   ⇒x^8 +1=Π_(k=0) ^7 (x−x_k )  ⇒J=∫(x^6 /(Π_(k=0) ^7 (x−x_k )))dx=∫Σ_(k=0) ^7 (a_k /(x−x_k ))dx  a_k =(x_k ^6 /(8x_k ^7 ))=(x_k ^7 /(−8)) ⇒ J=−(1/8)∫Σ_(k=0) ^7 ((x_k ^7 dx)/((x−x_k )))  ⇒J=−(1/8)Σ_(k=0) ^7 x_k ^7 ln(x−x_k )  ⇒∫(dx/(x^(10) +x^2 ))=−(1/x)+(1/8)Σ_(k=0) ^7 x_k ^7 ln(x−x_k )+C    Mr Mathmax′s favorite method. Haha!  I hope I′ve done it in the right way.

Commented by M±th+et+s last updated on 09/Jul/20

well done sir

Commented byAr Brandon last updated on 09/Jul/20

🍻

Commented bymathmax by abdo last updated on 09/Jul/20

yes your answer is correct

Commented byfloor(10²Eta[1]) last updated on 09/Jul/20

can you explain me how did you go from  here: ∫(x^6 /(Π_(k=0) ^7 (x−x_k )))dx to:=∫Σ_(k=0) ^7 (a_k /(x−x_k ))dx  and why a_k =(x_k ^6 /(8x_k ^7 ))

Commented byAr Brandon last updated on 09/Jul/20

Hi ! Mathmax explained this to me in Q86484.  I guess you have a look at it.

Commented byAr Brandon last updated on 09/Jul/20

  its a like theorem if F =((p(x))/(Q(x)))  with degp<degQ and  Q without real roots (⇒Q(x) =λΠ_i (x−z_i )) so  F(x)=Σ_i   (a_i /(x−z_i ))  and a_i =((p(z_i ))/(Q^′ (z_i )))

Commented byfloor(10²Eta[1]) last updated on 09/Jul/20

thanks! that′s really nice, do you know where   i can see the demonstration of this?

Commented byAr Brandon last updated on 09/Jul/20

Thanks for your approval Sir Mathmax. 😇

Commented byAr Brandon last updated on 09/Jul/20

No idea, I don't know where you can find this demo. Maybe a little search on Google might help.😃

Commented bymathmax by abdo last updated on 10/Jul/20

you are welcome sir