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Integral of $fracln(1+x^2)1+x^2$

How to find $int_0^inftyfracdx(1+x^2)^4}$Evaluating the integral $int_0^1arctan(1-x+x^2)dx$Closed-form expression of a definite integralHow do I find the following definite integral?How to integrate $frac{x^2log sin x1+x^6$Setting up this integral?Calculate $int_0^1fraccos(arctan x)sqrtxdx$Evaluate $intlimits_0^1fraclog(1-x+x^2)log(1+x-x^2)xdx$What is the simplest technique to evaluate the following definite triple integral?Indefinite integral $int arctan^2 x dx$ in terms of the dilogarithm function

1

$begingroup$

Compute $intfracln(1+x^2)1+x^2dx$

Attempt:

I tried to do Integration by Parts

$$u=ln(1+x^2)Rightarrow du=frac2xdx1+x^2\
dv=fracdx1+x^2Rightarrow v=arctan x\
intfracln(1+x^2)1+x^2dx=ln(1+x^2)arctan x-intfrac2xarctan x dx1+x^2$$

Since it seems that it not done in terms of elementary functions, what about the definite version:

$$int_0^1fracln(1+x^2)1+x^2dx$$

integration definite-integrals

share|cite|improve this question

edited Mar 14 at 18:15

Paras Khosla

2,348222

asked Mar 14 at 16:04

candcand

1,51911023

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The solution containes the dilog-function
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you possibly be interested in a definite version of this?
  $endgroup$
  – Math-fun
  Mar 14 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so aparent the indefinite version inst doable in term of elementary functions? What about the definite integral with interval from $0$ to $1$ or like $r$ to $1$ where $0<r<1$.
  $endgroup$
  – cand
  Mar 14 at 16:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is what Maple have done:$$i/2ln left( -i/2 left( x+i right) right) ln left( x-i right) +i/4 left( ln left( x-i right) right) ^2-i/2ln left( x-i right) ln left( x^2+1 right) +i/2it dilog left( -i/2 left( x+i right) right) -i/4 left( ln left( x+i right) right) ^2-i/2ln left( x+i right) ln left( i/2 left( x-i right) right) +i/2ln left( x+i right) ln left( x^2+1 right) -i/2it dilog left( i/2 left( x-i right) right) $$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 16:26
  

  
  
  
  

add a comment | 

1

$begingroup$

Compute $intfracln(1+x^2)1+x^2dx$

Attempt:

I tried to do Integration by Parts

$$u=ln(1+x^2)Rightarrow du=frac2xdx1+x^2\
dv=fracdx1+x^2Rightarrow v=arctan x\
intfracln(1+x^2)1+x^2dx=ln(1+x^2)arctan x-intfrac2xarctan x dx1+x^2$$

Since it seems that it not done in terms of elementary functions, what about the definite version:

$$int_0^1fracln(1+x^2)1+x^2dx$$

integration definite-integrals

share|cite|improve this question

edited Mar 14 at 18:15

Paras Khosla

2,348222

asked Mar 14 at 16:04

candcand

1,51911023

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The solution containes the dilog-function
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 16:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you possibly be interested in a definite version of this?
  $endgroup$
  – Math-fun
  Mar 14 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so aparent the indefinite version inst doable in term of elementary functions? What about the definite integral with interval from $0$ to $1$ or like $r$ to $1$ where $0<r<1$.
  $endgroup$
  – cand
  Mar 14 at 16:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That is what Maple have done:$$i/2ln left( -i/2 left( x+i right) right) ln left( x-i right) +i/4 left( ln left( x-i right) right) ^2-i/2ln left( x-i right) ln left( x^2+1 right) +i/2it dilog left( -i/2 left( x+i right) right) -i/4 left( ln left( x+i right) right) ^2-i/2ln left( x+i right) ln left( i/2 left( x-i right) right) +i/2ln left( x+i right) ln left( x^2+1 right) -i/2it dilog left( i/2 left( x-i right) right) $$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 16:26
  

  
  
  
  

add a comment | 

$begingroup$

Compute $intfracln(1+x^2)1+x^2dx$

Attempt:

I tried to do Integration by Parts

$$u=ln(1+x^2)Rightarrow du=frac2xdx1+x^2\
dv=fracdx1+x^2Rightarrow v=arctan x\
intfracln(1+x^2)1+x^2dx=ln(1+x^2)arctan x-intfrac2xarctan x dx1+x^2$$

Since it seems that it not done in terms of elementary functions, what about the definite version:

$$int_0^1fracln(1+x^2)1+x^2dx$$

integration definite-integrals

share|cite|improve this question

edited Mar 14 at 18:15

Paras Khosla

2,348222

asked Mar 14 at 16:04

candcand

1,51911023

$endgroup$

share|cite|improve this question

edited Mar 14 at 18:15

Paras Khosla

2,348222

asked Mar 14 at 16:04

candcand

1,51911023

share|cite|improve this question

edited Mar 14 at 18:15

Paras Khosla

2,348222

asked Mar 14 at 16:04

candcand

1,51911023

edited Mar 14 at 18:15

Paras Khosla

2,348222

edited Mar 14 at 18:15

Paras Khosla

2,348222

asked Mar 14 at 16:04

candcand

1,51911023

asked Mar 14 at 16:04

candcand

1,51911023

1 Answer
1

active

oldest

votes

1

$begingroup$

$$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$

---

Solving $2/icdot I_1$:
$$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$

---

Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$

---

Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$

share|cite|improve this answer

answered Mar 14 at 17:24

Paras KhoslaParas Khosla

2,348222

$endgroup$

add a comment | 

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1

$begingroup$

$$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$

---

Solving $2/icdot I_1$:
$$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$

---

Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$

---

Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$

share|cite|improve this answer

answered Mar 14 at 17:24

Paras KhoslaParas Khosla

2,348222

$endgroup$

add a comment | 

1

$begingroup$

$$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$

---

Solving $2/icdot I_1$:
$$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$

---

Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$

---

Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$

share|cite|improve this answer

answered Mar 14 at 17:24

Paras KhoslaParas Khosla

2,348222

$endgroup$

add a comment | 

$begingroup$

$$textLet I=intdfracln(1+x^2)1+x^2mathrm dx=intleft(underbracedfraciln(1+x^2)2(x+i)_I_1-underbracedfraciln(1+x^2)2(x-i)_I_2right)mathrm dx$$

---

Solving $2/icdot I_1$:
$$beginbmatrixu \ mathrm duendbmatrix=beginbmatrixx+i\ mathrm dxendbmatrix$$ $$beginalignintdfracln(1+x^2)x+imathrm dx&=intdfracln((u-i)^2+1)umathrm du\ & =intdfracln(u-2i)umathrm du+dfrac12ln^2(u)\ & = intdfracln(iu/2+1)umathrm du+ln(-2i)ln u+dfrac12ln^2u \ &= -mathrmLi_2left(-dfraciu2right)+dfrac12ln^2u+ln(-2i)ln(u)\&=-mathrmLi_2left(-dfraci(x+i)2right)+dfrac12 ln^2(x+i)+ln(-2i)ln(x+i)tag1endalign$$

---

Similarly solve $2/icdot I_2$ to get the following result: $$intdfracln(1+x^2)x-imathrm dx = -mathrmLi_2left(-dfraci(x-i)2right)+dfrac12ln^2(x-i)+ln(2i)ln(x-i)tag2$$

---

Computing $i/2cdot [(1)+(2)]equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=dfraci4left[lnmid x+i mid left(lnmid x+imid+2ln(-2i)right)-lnmid x-imidleft(lnmid x-imid +2ln(2i)right)\ +2mathrmLi_2left(dfracix+12right)-2mathrmLi_2left(-dfracix-12right)right]+C$$

share|cite|improve this answer

answered Mar 14 at 17:24

Paras KhoslaParas Khosla

2,348222

$endgroup$

share|cite|improve this answer

answered Mar 14 at 17:24

Paras KhoslaParas Khosla

2,348222

answered Mar 14 at 17:24

Paras KhoslaParas Khosla

2,348222

answered Mar 14 at 17:24

Paras KhoslaParas Khosla

2,348222