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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

my post here never get answers y??

$${my}\:{post}\:{here}\:{never}\: \\ $$$${get}\:{answers}\:{y}?? \\ $$

Commented by mr W last updated on 07/May/20

it′s not due you personally, but mainly due to your question, maybe the question is just too hard for the people here.

$${it}'{s}\:{not}\:{due}\:{you}\:{personally},\:{but}\:{mainly} \\ $$$${due}\:{to}\:{your}\:{question},\:{maybe}\:{the} \\ $$$${question}\:{is}\:{just}\:{too}\:{hard}\:{for}\:{the} \\ $$$${people}\:{here}. \\ $$

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

what′s your logic sir? you need help doesn′t mean that other people can and should help you. as i have said, maybe your question is too hard, other people want to, but are not able to help you.

$${what}'{s}\:{your}\:{logic}\:{sir}?\:{you}\:{need}\:{help} \\ $$$${doesn}'{t}\:{mean}\:{that}\:{other}\:{people}\:{can} \\ $$$${and}\:{should}\:{help}\:{you}.\:\:{as}\:{i}\:{have}\:{said}, \\ $$$${maybe}\:{your}\:{question}\:{is}\:{too}\:{hard},\:{other} \\ $$$${people}\:{want}\:{to},\:{but}\:{are}\:{not}\:{able}\:{to} \\ $$$${help}\:{you}. \\ $$

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

$${x}^{\mathrm{4}} +\mathrm{1}=\left({x}^{\mathrm{2}} −{i}\right)\left({x}^{\mathrm{2}} +{i}\right) \\ $$$$=\left({x}−\sqrt{{i}}\right)\left({x}+\sqrt{{i}}\right)\left({x}−{i}\sqrt{{i}}\right)\left({x}+{i}\sqrt{{i}}\right) \\ $$$${x}^{\mathrm{4}} +\mathrm{1}\:\mathrm{has}\:\mathrm{4}\:\mathrm{roots}\:\left(\mathrm{4}^{\mathrm{th}} \mathrm{root}\:\mathrm{of}\:\mathrm{umiy}\right) \\ $$$$\mathrm{For}\:\mathrm{ease}\:\mathrm{in}\:\mathrm{typing}\:\mathrm{let}\:\mathrm{us}\:\mathrm{write}\:\mathrm{that}\:\mathrm{as} \\ $$$${a},−{a},{b},−{b} \\ $$$$\frac{\mathrm{1}}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\left({x}+{a}\right)^{\mathrm{2}} \left({x}−{a}\right)^{\mathrm{2}} \left({x}+{b}\right)^{\mathrm{2}} \left({x}−{b}\right)^{\mathrm{2}} } \\ $$$${if}\:{we}\:{use}\:{partial}\:{fraction}\:{you}\:{will} \\ $$$${get}\:{following}\:{form} \\ $$$$\frac{{A}}{\left({x}−{a}\right)}+\frac{{B}}{\left({x}−{a}\right)^{\mathrm{2}} }+...\: \\ $$$${basically}\:\frac{{constant}}{{linear}}+\frac{{constant}}{{quadratic}} \\ $$$${with}\:{these}\:{step}\: \\ $$$${the}\:{final}\:\mathrm{8}\:{integrals}\:{will}\:{be}\:{of}\:{two} \\ $$$${forms}\: \\ $$$$\frac{{Cx}}{\left({x}+{k}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}},\:\frac{{Cx}}{\left({x}+{k}\right)\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}} \\ $$$$={which}\:{can}\:{be}\:{integrated}\:{using} \\ $$$${standard}\:{formulas} \\ $$

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

$$\mathrm{You}\:\mathrm{will}\:\mathrm{get}\:\mathrm{very}\:\mathrm{long}\:\mathrm{answer}\:\mathrm{value}\:\mathrm{for}\:\mathrm{integral} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{source}\:\mathrm{of}\:\mathrm{question} \\ $$

Commented by MJS last updated on 07/May/20

I tried almost everything I know but I cannot solve this integral...

$$\mathrm{I}\:\mathrm{tried}\:\mathrm{almost}\:\mathrm{everything}\:\mathrm{I}\:\mathrm{know}\:\mathrm{but}\:\mathrm{I}\:\mathrm{cannot} \\ $$$$\mathrm{solve}\:\mathrm{this}\:\mathrm{integral}... \\ $$

Commented by turbo msup by abdo last updated on 08/May/20

i will post snswer by i cant complet it

$${i}\:{will}\:{post}\:{snswer}\:{by}\:{i}\:{cant}\:{complet} \\ $$$${it} \\ $$

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...

$${let}\:{f}\left({a}\right)\:=\int\:\:\frac{{xdx}}{\left({x}^{\mathrm{4}} \:+{a}^{\mathrm{4}} \right)\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:\:\:\left({a}>\mathrm{0}\right) \\ $$$${we}\:{have}\:{f}^{'} \left({a}\right)\:=−\int\:\:\:\frac{\mathrm{4}{a}^{\mathrm{3}} {xdx}}{\left({x}^{\mathrm{4}} +{a}^{\mathrm{4}} \right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:\Rightarrow \\ $$$$\int\:\:\:\frac{{xdx}}{\left({x}^{\mathrm{4}} \:+{a}^{\mathrm{4}} \right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:=−\frac{{f}^{'} \left({a}\right)}{\mathrm{4}{a}^{\mathrm{3}} }\:\:{let}\:{explicit}\:{f}\left({a}\right) \\ $$$${f}\left({a}\right)\:=_{{x}={at}} \:\:\:\:\int\:\:\:\frac{{at}\:{adt}}{{a}^{\mathrm{4}} \left({t}^{\mathrm{4}} +\mathrm{1}\right)\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} −{at}\:+\mathrm{1}}} \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\int\:\:\:\frac{{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)\sqrt{\left({at}\right)^{\mathrm{2}} −\mathrm{2}\frac{{at}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\int\:\:\:\frac{{tdt}}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)\sqrt{\left({at}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$=_{{at}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({u}\right)} \:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} }×\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\int\:\:\:\frac{\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\mathrm{1}+\sqrt{\mathrm{3}}{chu}\right)×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}{a}}{sh}\left({u}\right){du}}{{shu}\:\left(\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\mathrm{1}+\sqrt{\mathrm{3}}{chu}\right)\right)^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{16}}{{a}^{\mathrm{4}} }\:\int\:\:\:\:\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}{chu}\right){du}}{\left(\mathrm{1}+\sqrt{\mathrm{3}}{chu}\right)^{\mathrm{4}} }\:=\frac{\mathrm{16}}{{a}^{\mathrm{4}} }\int\:\:\:\frac{{du}}{\left(\sqrt{\mathrm{3}}{chu}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{16}}{{a}^{\mathrm{4}} }\:\int\:\:\:\:\frac{{du}}{\mathrm{3}\sqrt{\mathrm{3}}{ch}^{\mathrm{3}} {u}\:\:+\mathrm{9}\:{ch}^{\mathrm{2}} {u}\:\:+\mathrm{3}\sqrt{\mathrm{3}}{chu}\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{16}}{{a}^{\mathrm{4}} }\int\:\:\frac{{du}}{\mathrm{3}\sqrt{\mathrm{3}}\left(\frac{{e}^{{u}} +{e}^{−{u}} }{\mathrm{2}}\right)^{\mathrm{3}} \:+\mathrm{9}\left(\frac{{e}^{{u}} +{e}^{−{u}} }{\mathrm{2}}\right)^{\mathrm{2}} \:+\mathrm{3}\sqrt{\mathrm{3}}×\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{16}}{{a}^{\mathrm{4}} }\:\int\:\:\:\frac{{du}}{\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\left({e}^{{u}} \:+{e}^{−{u}} \right)^{\mathrm{3}} \:+\frac{\mathrm{9}}{\mathrm{4}}\left({e}^{{u}} \:+{e}^{−{u}} \right)^{\mathrm{2}} \:+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\left({e}^{{u}} \:+{e}^{−{u}} \right)+\mathrm{1}} \\ $$$$=\frac{\mathrm{16}×\mathrm{8}}{{a}^{\mathrm{4}} }\:\int\:\:\:\:\frac{{du}}{\mathrm{3}\sqrt{\mathrm{3}}\left({e}^{{u}} \:+{e}^{−{u}} \right)^{\mathrm{3}} \:+\mathrm{18}\left({e}^{{u}} \:+{e}^{−{u}} \right)^{\mathrm{2}} \:+\mathrm{12}\sqrt{\mathrm{3}}\left({e}^{{u}} \:+{e}^{−{u}} \right)+\mathrm{8}} \\ $$$$=_{{e}^{{u}} ={z}} \:\:\:\frac{\mathrm{128}}{{a}^{\mathrm{4}} }\:\:\int\:\:\:\:\:\frac{{dz}}{{z}\left\{\mathrm{3}\sqrt{\mathrm{3}}\left({z}+{z}^{−\mathrm{1}} \right)^{\mathrm{3}} \:+\mathrm{18}\left({z}+{z}^{−\mathrm{1}} \right)^{\mathrm{2}} \:+\mathrm{12}\sqrt{\mathrm{3}}\left({z}+{z}^{−\mathrm{1}} \right)\:+\mathrm{8}\right.} \\ $$$$....{be}\:{continued}... \\ $$$$ \\ $$

Commented by  M±th+et+s last updated on 08/May/20

its nice to see you again in the forum sir prakash jain

$${its}\:{nice}\:{to}\:{see}\:{you}\:{again}\:{in}\:{the}\:{forum} \\ $$$${sir}\:{prakash}\:{jain} \\ $$

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