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Question Number 87279 by Ar Brandon last updated on 03/Apr/20
∫(x^2 /(1+x^4 ))dx
$$\int\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$
Commented by abdomathmax last updated on 03/Apr/20
complex method let decompose F(x)=(x^2 /(x^4 +1)) x^4 +1=0 ⇒x^4 =−1 =e^(i(2k+1)π) ⇒x_k =e^(i(((2k+1)π)/4)) and k∈[[0,3]] so the roots are x_0 =e^((iπ)/4) ,x_1 =e^((i3π)/4) , x_2 =e^(i((5π)/4)) ,x_3 = e^(i((7π)/4)) and F(x) =Σ_(k=0) ^3 (a_k /(x−x_k )) a_k =(x_k ^2 /(4x_k ^3 )) =−(1/4)x_k ^3 ⇒F(x) =−(1/4)Σ_(k=0) ^3 (x_k ^3 /(x−x_k )) ⇒ ∫ (x^2 /(1+x^4 ))dx =−(1/4)Σ_(k=0) ^3 x_k ^3 ln(x−x_k ) + C
$${complex}\:{method}\:\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$${x}^{\mathrm{4}} +\mathrm{1}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{4}} =−\mathrm{1}\:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow{x}_{{k}} ={e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{4}}} \\ $$$${and}\:{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right]\:{so}\:{the}\:{roots}\:\:{are} \\ $$$${x}_{\mathrm{0}} ={e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:,{x}_{\mathrm{1}} ={e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \:,\:\:{x}_{\mathrm{2}} ={e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \:\:\:,{x}_{\mathrm{3}} =\:{e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{4}}} \:{and} \\ $$$${F}\left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} \:\frac{{a}_{{k}} }{{x}−{x}_{{k}} } \\ $$$${a}_{{k}} =\frac{{x}_{{k}} ^{\mathrm{2}} }{\mathrm{4}{x}_{{k}} ^{\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{4}}{x}_{{k}} ^{\mathrm{3}} \:\Rightarrow{F}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} \frac{{x}_{{k}} ^{\mathrm{3}} }{{x}−{x}_{{k}} }\:\Rightarrow \\ $$$$\int\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:=−\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} {x}_{{k}} ^{\mathrm{3}} {ln}\left({x}−{x}_{{k}} \right)\:+\:{C} \\ $$
Commented by Ar Brandon last updated on 04/Apr/20
Please may I know how you switched from a_k =(x_k ^2 /(4x_k ^3 )) to a_k =−(1/4)x_k ^3 ?
$${Please}\:\:{may}\:\:{I}\:\:{know}\:\:{how}\:\:{you}\:\:{switched}\:\:{from}\:\: \\ $$$${a}_{{k}} =\frac{{x}_{{k}} ^{\mathrm{2}} }{\mathrm{4}{x}_{{k}} ^{\mathrm{3}} }\:\:{to}\:\:{a}_{{k}} =−\frac{\mathrm{1}}{\mathrm{4}}{x}_{{k}} ^{\mathrm{3}} \:\:\:? \\ $$
Answered by TANMAY PANACEA. last updated on 03/Apr/20
∫(dx/(x^2 +(1/x^2 ))) (1/2)∫((1−(1/x^2 )+1+(1/x^2 ))/(x^2 +(1/x^2 )))dx (1/2)∫((d(x+(1/x)))/((x+(1/x))^2 −2))+(1/2)∫((d(x−(1/x)))/((x−(1/x))^2 +2)) (1/2)×(1/(2(√2)))ln(((x+(1/x)−(√2))/(x+(1/x)+(√2))))+(1/2)×(1/( (√2)))tan^(−1) (((x−(1/x))/( (√2))))+c
$$\int\frac{{dx}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$
Commented by Ar Brandon last updated on 03/Apr/20
Right
$${Right} \\ $$
Commented by TANMAY PANACEA. last updated on 03/Apr/20
thank hou sir
$${thank}\:{hou}\:{sir} \\ $$
Commented by peter frank last updated on 03/Apr/20
thank you
$${thank}\:{you} \\ $$
Answered by redmiiuser last updated on 03/Apr/20
x=(√(tanz)) dx=((sec^2 z.dz)/(2.(√(tan z)))) ∫((tan z.sec^2 z.dz)/(2.(√(tan z)).sec^2 z)) =(1/2)∫(√(tan z)).dz sec z=u du=sec z.tan z.dz (1/2)∫((√(tan z))/(sec z.tan z))du =(1/2)∫(du/(u.(√(u^2 −1)))) =(1/2)∫(((−1)^((−1)/2) .(1−u^2 )^((−1)/2) )/u)du =(((−1)^((−1)/2) )/2)Σ_(n=0) ^∞ ((2n!.u^(2n) )/(2^(2n) .(n!)^2 .(2n))) =(((−1)^((−1)/2) )/2)Σ_(n=0) ^∞ ((2n!.(1+x^4 )^n )/(2^(2n) .(n!)^2 .2n))
$${x}=\sqrt{{tanz}} \\ $$$${dx}=\frac{\mathrm{sec}\:^{\mathrm{2}} {z}.{dz}}{\mathrm{2}.\sqrt{\mathrm{tan}\:{z}}} \\ $$$$\int\frac{\mathrm{tan}\:{z}.\mathrm{sec}\:^{\mathrm{2}} {z}.{dz}}{\mathrm{2}.\sqrt{\mathrm{tan}\:{z}}.\mathrm{sec}\:^{\mathrm{2}} {z}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\mathrm{tan}\:{z}}.{dz} \\ $$$$\mathrm{sec}\:{z}={u} \\ $$$${du}=\mathrm{sec}\:{z}.\mathrm{tan}\:{z}.{dz} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{\mathrm{tan}\:{z}}}{\mathrm{sec}\:{z}.\mathrm{tan}\:{z}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}.\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(−\mathrm{1}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} .\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{{u}}{du} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}!.{u}^{\mathrm{2}{n}} }{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} .\left(\mathrm{2}{n}\right)} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}!.\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{{n}} }{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} .\mathrm{2}{n}} \\ $$$$ \\ $$$$ \\ $$
Commented by redmiiuser last updated on 03/Apr/20
pls check!
$${pls}\:{check}! \\ $$
Commented by redmiiuser last updated on 03/Apr/20
Mr.Ar Brandon I request you to check my solution.
$${Mr}.{Ar}\:{Brandon}\:{I} \\ $$$${request}\:{you}\:{to}\:{check} \\ $$$${my}\:{solution}. \\ $$
Commented by MJS last updated on 03/Apr/20
Mr. this forum had been a polite place before you arrived. nobody will answer you if you don′t change your behaviour.
$$\mathrm{Mr}.\:\mathrm{this}\:\mathrm{forum}\:\mathrm{had}\:\mathrm{been}\:\mathrm{a}\:\mathrm{polite}\:\mathrm{place}\:\mathrm{before} \\ $$$$\mathrm{you}\:\mathrm{arrived}.\:\mathrm{nobody}\:\mathrm{will}\:\mathrm{answer}\:\mathrm{you}\:\mathrm{if}\:\mathrm{you} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{change}\:\mathrm{your}\:\mathrm{behaviour}. \\ $$
Commented by mr W last updated on 03/Apr/20
100% agree!
$$\mathrm{100\%}\:{agree}! \\ $$
Commented by mr W last updated on 03/Apr/20
i would say: if you are sure that you can solve a question, then solve it! people will see it and give comment when they can and when they like. to force other people to check it or to give comment is not polite, not cool, even boring! and if you can not solve a question or you are not sure with your solution, you can also leave it. nobody will blame you if you don′t give answer to every question.
$${i}\:{would}\:{say}:\:{if}\:{you}\:{are}\:{sure}\:{that}\:{you} \\ $$$${can}\:{solve}\:{a}\:{question},\:{then}\:{solve}\:{it}! \\ $$$${people}\:{will}\:{see}\:{it}\:{and}\:{give}\:{comment} \\ $$$${when}\:{they}\:{can}\:{and}\:{when}\:{they}\:{like}. \\ $$$${to}\:{force}\:{other}\:{people}\:{to}\:{check}\:{it}\:{or} \\ $$$${to}\:{give}\:{comment}\:{is}\:{not}\:{polite},\:{not} \\ $$$${cool},\:{even}\:{boring}!\:{and}\:{if}\:{you}\:{can}\:{not}\: \\ $$$${solve}\:{a}\:{question}\:{or}\:{you}\:{are}\:{not}\:{sure} \\ $$$${with}\:{your}\:{solution},\:{you}\:{can}\:{also} \\ $$$${leave}\:{it}.\:{nobody}\:{will}\:{blame}\:{you}\:{if}\:{you} \\ $$$${don}'{t}\:{give}\:{answer}\:{to}\:{every}\:{question}. \\ $$
Commented by Ar Brandon last updated on 04/Apr/20
Don′t get it.
$${Don}'{t}\:\:{get}\:\:{it}. \\ $$
Commented by ajfour last updated on 04/Apr/20
i think these suggestions are meant for mr redmuiser..
$${i}\:{think}\:{these}\:{suggestions}\:{are} \\ $$$${meant}\:{for}\:{mr}\:{redmuiser}.. \\ $$
Commented by MJS last updated on 04/Apr/20
yes
$$\mathrm{yes} \\ $$
Commented by redmiiuser last updated on 04/Apr/20
ok thanks everyone for such beautiful comments.I urged only to check and I didn′t forced to comment.
$${ok}\:{thanks}\:{everyone} \\ $$$${for}\:{such}\:{beautiful} \\ $$$${comments}.{I}\:{urged}\:{only} \\ $$$${to}\:{check}\:{and}\:{I}\:{didn}'{t}\: \\ $$$${forced}\:{to}\:{comment}. \\ $$
Commented by redmiiuser last updated on 04/Apr/20
And also one who gives question must check every answer. Its his/her duty I only asked to check my solution and in many questions people ask to check the answer and there is no point to insult anyone. Anyway thanks mr.MJS , mr.W to give such comments.
$${And}\:{also}\:{one}\:{who} \\ $$$${gives}\:{question}\:{must} \\ $$$${check}\:{every}\:{answer}. \\ $$$${Its}\:{his}/{her}\:{duty}\:{I} \\ $$$${only}\:{asked}\:{to}\:{check} \\ $$$${my}\:{solution}\:{and}\:{in} \\ $$$${many}\:{questions}\:{people} \\ $$$${ask}\:{to}\:{check}\:{the}\:{answer} \\ $$$${and}\:{there}\:{is}\:{no}\:{point} \\ $$$${to}\:{insult}\:{anyone}. \\ $$$${Anyway}\:{thanks} \\ $$$${mr}.{MJS}\:,\:{mr}.{W}\:{to}\:{give} \\ $$$${such}\:{comments}. \\ $$$$ \\ $$
Commented by redmiiuser last updated on 04/Apr/20
If for me this forum is not getting a polite place then please excuse me its my tendency to get excited.I am sorry Honourable teachers.
$${If}\:{for}\:{me}\:{this}\:{forum} \\ $$$${is}\:{not}\:\:{getting}\:{a}\:{polite}\:{place} \\ $$$${then}\:{please}\:{excuse} \\ $$$${me}\:{its}\:{my}\:{tendency} \\ $$$${to}\:{get}\:{excited}.{I}\:{am}\: \\ $$$${sorry}\:{Honourable} \\ $$$${teachers}. \\ $$
Commented by Ar Brandon last updated on 04/Apr/20
I don′t get the method.
$${I}\:\:{don}'{t}\:\:{get}\:\:{the}\:\:{method}. \\ $$
Commented by redmiiuser last updated on 04/Apr/20
why mister can you tell me the part where you faced problem
$${why}\:{mister}\:{can}\:{you}\:{tell} \\ $$$${me}\:{the}\:{part}\:{where}\:{you} \\ $$$${faced}\:{problem} \\ $$
Commented by ajfour last updated on 05/Apr/20
Commented by redmiiuser last updated on 05/Apr/20
hahaha... nice sketch
$${hahaha}…\:{nice}\:{sketch} \\ $$
Commented by ajfour last updated on 05/Apr/20
thankx, was meant 4 u.
Commented by Ar Brandon last updated on 05/Apr/20
No worries. I′ll review it some other time. I realised you love using the summation series while solving integrals. Haha
$${No}\:{worries}.\:{I}'{ll}\:{review}\:{it}\:{some} \\ $$$${other}\:{time}.\: \\ $$$${I}\:{realised}\:{you}\:{love}\:{using}\:{the}\:{summation}\:{series} \\ $$$${while}\:{solving}\:{integrals}.\:{Haha} \\ $$
Answered by MJS last updated on 04/Apr/20
∫(x^2 /(x^4 +1))dx= =((√2)/4)∫(x/(x^2 −(√2)x+1))dx−((√2)/4)∫(x/(x^2 +(√2)x+1))dx= =((√2)/8)ln (x^2 −(√2)x+1) +((√2)/4)arctan ((√2)x−1) − −((√2)/8)ln (x^2 +(√2)x+1) +((√2)/4)arctan ((√2)x+1) = =((√2)/8)(ln ((x^2 −(√2)x+1)/(x^2 −(√2)x+1)) +2(arctan ((√2)x−1) +arctan ((√2)x+1))) +C
$$\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{1}}{dx}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{x}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{x}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:− \\ $$$$\:\:\:\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\left(\mathrm{ln}\:\frac{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+\mathrm{2}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\right)\right)\:+{C} \\ $$

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