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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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