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Question Number 92521 by hovero clinton last updated on 07/May/20

Commented by hovero clinton last updated on 07/May/20

$m y p o s t h e r e n e v e r$ $g e t a n s w e r s y ? ?$
Commented by mr W last updated on 07/May/20

${i t}^{'} s n o t d u e y o u p e r s o n a l l y, b u t m a i n l y$ $d u e t o y o u r q u e s t i o n, m a y b e t h e$ $q u e s t i o n i s j u s t t o o h a r d f o r t h e$ $p e o p l e h e r e .$
Commented by hovero clinton last updated on 07/May/20
ok sir but I post questions I need help in
Commented by mr W last updated on 07/May/20

${w h a t}^{'} s y o u r l o g i c s i r ? y o u n e e d h e l p$ ${d o e s n}^{'} t m e a n t h a t o t h e r p e o p l e c a n$ $a n d s h o u l d h e l p y o u . a s i h a v e s a i d,$ $m a y b e y o u r q u e s t i o n i s t o o h a r d, o t h e r$ $p e o p l e w a n t t o, b u t a r e n o t a b l e t o$ $h e l p y o u .$
Commented by prakash jain last updated on 07/May/20
$x^4 +1=(x^2 −i)(x^2 +i) =(x−(√i))(x+(√i))(x−i(√i))(x+i(√i)) x^4 +1 has 4 roots (4^(th) root of umiy) For ease in typing let us write that as a,−a,b,−b (1/((x^4 +1)^2 ))=(1/((x+a)^2 (x−a)^2 (x+b)^2 (x−b)^2 )) if we use partial fraction you will get following form (A/((x−a)))+(B/((x−a)^2 ))+... basically ((constant)/(linear))+((constant)/(quadratic)) with these step the final 8 integrals will be of two forms ((Cx)/((x+k)^2 (√(x^2 −x+1)))), ((Cx)/((x+k)(√(x^2 −x+1)))) =which can be integrated using standard formulas$
$x^{4} + 1 = (x^{2} - i) (x^{2} + i)$ $= (x - \sqrt{i}) (x + \sqrt{i}) (x - i \sqrt{i}) (x + i \sqrt{i})$ $x^{4} + 1 has 4 roots (4^{th} root of umiy)$ $For ease in typing let us write that as$ $a, - a, b, - b$ $\frac{1}{{(x^{4} + 1)}^{2}} = \frac{1}{{(x + a)}^{2} {(x - a)}^{2} {(x + b)}^{2} {(x - b)}^{2}}$ $i f w e u s e p a r t i a l f r a c t i o n y o u w i l l$ $g e t f o l l o w i n g f o r m$ $\frac{A}{(x - a)} + \frac{B}{{(x - a)}^{2}} + . . .$ $b a s i c a l l y \frac{c o n s t a n t}{l i n e a r} + \frac{c o n s t a n t}{q u a d r a t i c}$ $w i t h t h e s e s t e p$ $t h e f i n a l 8 i n t e g r a l s w i l l b e o f t w o$ $f o r m s$ $\frac{C x}{{(x + k)}^{2} \sqrt{x^{2} - x + 1}}, \frac{C x}{(x + k) \sqrt{x^{2} - x + 1}}$ $= w h i c h c a n b e i n t e g r a t e d u s i n g$ $s t a n d a r d f o r m u l a s$
Commented by prakash jain last updated on 07/May/20

$You will get very long answer value for integral$ $What is the source of question$
Commented by MJS last updated on 07/May/20

$I tried almost everything I know but I cannot$ $solve this integral . . .$
Commented by turbo msup by abdo last updated on 08/May/20

$i w i l l p o s t s n s w e r b y i c a n t c o m p l e t$ $i t$
Commented by abdomathmax last updated on 08/May/20
$let f(a) =∫ ((xdx)/((x^4 +a^4 )(√(x^2 −x+1)))) (a>0) we have f^′ (a) =−∫ ((4a^3 xdx)/((x^4 +a^4 )^2 (√(x^2 −x+1)))) ⇒ ∫ ((xdx)/((x^4 +a^4 )^2 (√(x^2 −x+1)))) =−((f^′ (a))/(4a^3 )) let explicit f(a) f(a) =_(x=at) ∫ ((at adt)/(a^4 (t^4 +1)(√(a^2 t^2 −at +1)))) =(1/a^2 )∫ ((tdt)/((1+t^4 )(√((at)^2 −2((at)/2)+(1/4)+(3/4))))) =(1/a^2 )∫ ((tdt)/((t^4 +1)(√((at−(1/2))^2 +(3/4))))) =_(at−(1/2)=((√3)/2)ch(u)) (1/a^2 )×(2/(√3)) ∫ (((1/(2a))(1+(√3)chu)×((√3)/(2a))sh(u)du)/(shu ((1/(2a))(1+(√3)chu))^4 )) =((16)/a^4 ) ∫ (((1+(√3)chu)du)/((1+(√3)chu)^4 )) =((16)/a^4 )∫ (du/(((√3)chu+1)^3 )) =((16)/a^4 ) ∫ (du/(3(√3)ch^3 u +9 ch^2 u +3(√3)chu +1)) =((16)/a^4 )∫ (du/(3(√3)(((e^u +e^(−u) )/2))^3 +9(((e^u +e^(−u) )/2))^2 +3(√3)×((e^u +e^(−u) )/2)+1)) =((16)/a^4 ) ∫ (du/(((3(√3))/8)(e^u +e^(−u) )^3 +(9/4)(e^u +e^(−u) )^2 +((3(√3))/2)(e^u +e^(−u) )+1)) =((16×8)/a^4 ) ∫ (du/(3(√3)(e^u +e^(−u) )^3 +18(e^u +e^(−u) )^2 +12(√3)(e^u +e^(−u) )+8)) =_(e^u =z) ((128)/a^4 ) ∫ (dz/(z{3(√3)(z+z^(−1) )^3 +18(z+z^(−1) )^2 +12(√3)(z+z^(−1) ) +8)) ....be continued...$
$l e t f (a) = \int \frac{x d x}{(x^{4} + a^{4}) \sqrt{x^{2} - x + 1}} (a > 0)$ $w e h a v e f^{^{'}} (a) = - \int \frac{4 a^{3} x d x}{{(x^{4} + a^{4})}^{2} \sqrt{x^{2} - x + 1}} \Rightarrow$ $\int \frac{x d x}{{(x^{4} + a^{4})}^{2} \sqrt{x^{2} - x + 1}} = - \frac{f^{^{'}} (a)}{4 a^{3}} l e t e x p l i c i t f (a)$ $f (a) =_{x = a t} \int \frac{a t a d t}{a^{4} (t^{4} + 1) \sqrt{a^{2} t^{2} - a t + 1}}$ $= \frac{1}{a^{2}} \int \frac{t d t}{(1 + t^{4}) \sqrt{{(a t)}^{2} - 2 \frac{a t}{2} + \frac{1}{4} + \frac{3}{4}}}$ $= \frac{1}{a^{2}} \int \frac{t d t}{(t^{4} + 1) \sqrt{{(a t - \frac{1}{2})}^{2} + \frac{3}{4}}}$ $=_{a t - \frac{1}{2} = \frac{\sqrt{3}}{2} c h (u)} \frac{1}{a^{2}} \times \frac{2}{\sqrt{3}} \int \frac{\frac{1}{2 a} (1 + \sqrt{3} c h u) \times \frac{\sqrt{3}}{2 a} s h (u) d u}{s h u {(\frac{1}{2 a} (1 + \sqrt{3} c h u))}^{4}}$ $= \frac{16}{a^{4}} \int \frac{(1 + \sqrt{3} c h u) d u}{{(1 + \sqrt{3} c h u)}^{4}} = \frac{16}{a^{4}} \int \frac{d u}{{(\sqrt{3} c h u + 1)}^{3}}$ $= \frac{16}{a^{4}} \int \frac{d u}{3 \sqrt{3} {c h}^{3} u + 9 {c h}^{2} u + 3 \sqrt{3} c h u + 1}$ $= \frac{16}{a^{4}} \int \frac{d u}{3 \sqrt{3} {(\frac{e^{u} + e^{- u}}{2})}^{3} + 9 {(\frac{e^{u} + e^{- u}}{2})}^{2} + 3 \sqrt{3} \times \frac{e^{u} + e^{- u}}{2} + 1}$ $= \frac{16}{a^{4}} \int \frac{d u}{\frac{3 \sqrt{3}}{8} {(e^{u} + e^{- u})}^{3} + \frac{9}{4} {(e^{u} + e^{- u})}^{2} + \frac{3 \sqrt{3}}{2} (e^{u} + e^{- u}) + 1}$ $= \frac{16 \times 8}{a^{4}} \int \frac{d u}{3 \sqrt{3} {(e^{u} + e^{- u})}^{3} + 18 {(e^{u} + e^{- u})}^{2} + 12 \sqrt{3} (e^{u} + e^{- u}) + 8}$ $=_{e^{u} = z} \frac{128}{a^{4}} \int \frac{d z}{z {3 \sqrt{3} {(z + z^{- 1})}^{3} + 18 {(z + z^{- 1})}^{2} + 12 \sqrt{3} (z + z^{- 1}) + 8}$ $. . . . b e c o n t i n u e d . . .$
Commented by M±th+et+s last updated on 08/May/20

$i t s n i c e t o s e e y o u a g a i n i n t h e f o r u m$ $s i r p r a k a s h j a i n$
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