Integral with inverse trigonometric function [closed]

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How do I integrate $$int cot^-1sqrtx^2+x+1 dx$$
I don't understand how to proceed?
I did try the question using integration by parts, but it didn’t help.

integration

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edited Aug 6 at 16:51

Dylan

11.4k31026

asked Aug 6 at 12:52

Sidharth Giri

133

closed as off-topic by amWhy, José Carlos Santos, Adrian Keister, Manthanein, Nosrati Aug 7 at 7:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Adrian Keister, Manthanein, Nosrati

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a serious mess.
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  Aug 6 at 12:56
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

1

How do I integrate $$int cot^-1sqrtx^2+x+1 dx$$
I don't understand how to proceed?
I did try the question using integration by parts, but it didn’t help.

integration

share|cite|improve this question

edited Aug 6 at 16:51

Dylan

11.4k31026

asked Aug 6 at 12:52

Sidharth Giri

133

closed as off-topic by amWhy, José Carlos Santos, Adrian Keister, Manthanein, Nosrati Aug 7 at 7:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Adrian Keister, Manthanein, Nosrati

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a serious mess.
  – lulu
  Aug 6 at 12:56
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

How do I integrate $$int cot^-1sqrtx^2+x+1 dx$$
I don't understand how to proceed?
I did try the question using integration by parts, but it didn’t help.

integration

share|cite|improve this question

edited Aug 6 at 16:51

Dylan

11.4k31026

asked Aug 6 at 12:52

Sidharth Giri

133

How do I integrate $$int cot^-1sqrtx^2+x+1 dx$$
I don't understand how to proceed?
I did try the question using integration by parts, but it didn’t help.

integration

share|cite|improve this question

edited Aug 6 at 16:51

Dylan

11.4k31026

asked Aug 6 at 12:52

Sidharth Giri

133

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 16:51

Dylan

11.4k31026

edited Aug 6 at 16:51

Dylan

11.4k31026

edited Aug 6 at 16:51

Dylan

11.4k31026

11.4k31026

asked Aug 6 at 12:52

Sidharth Giri

133

asked Aug 6 at 12:52

Sidharth Giri

133

asked Aug 6 at 12:52

Sidharth Giri

133

133

closed as off-topic by amWhy, José Carlos Santos, Adrian Keister, Manthanein, Nosrati Aug 7 at 7:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Adrian Keister, Manthanein, Nosrati

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, José Carlos Santos, Adrian Keister, Manthanein, Nosrati Aug 7 at 7:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Adrian Keister, Manthanein, Nosrati

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a serious mess.
  – lulu
  Aug 6 at 12:56
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's a serious mess.
  – lulu
  Aug 6 at 12:56
  

  
  
  
  

1

1

It's a serious mess.
– lulu
Aug 6 at 12:56

It's a serious mess.
– lulu
Aug 6 at 12:56

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

A solution without using hyperbolic functions

Integration by parts gives
$$xcot^-1left( sqrtx^2+x+1 right)+frac12int fracx(2x+1)sqrtx^2+x+1(x^2+x+2),dx$$
i can split the integrated on the right as
$$
beginalign
& =2intfrac1sqrtx^2+x+1dx-intfrac1+2x2sqrtx^2+x+1left( x^2+x+2 right)dx-frac72intfrac1sqrt1+x+x^2left( x^2+x+2 right)dx \
& =2A-B-frac72C \
endalign
$$
to calculate $A$ notice that $x^2+x+1=left( x+frac12 right)^2+left( fracsqrt32 right)^2$ and using $x+frac12=fracsqrt32u$

$$A=intfrac1sqrtu^2+1du=ln left( u+sqrtu^2+1 right)=ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)$$
that’s because i accept the fact that $left( ln left( u+sqrtu^2+1 right) right)^prime =frac1sqrtu^2+1$

for $B$ we have
$$frac1+2x2sqrt1+x+x^2left( 2+x+x^2 right)=fracleft( sqrt1+x+x^2 right)^prime left( 1+left( sqrt1+x+x^2 right)^2 right)Rightarrow B=tan ^-1left( sqrt1+x+x^2 right)$$

for $C$ use $u=frac2x+1sqrtx^2+x+1$ hence $du=frac32left( sqrtx^2+x+1 right)^3dx$ also
$$u^2-7=frac4x^2+4x+1x^2+x+1-frac7x^2+7x+7x^2+x+1=frac-3x^2-3x-6x^2+x+1=-3left( fracx^2+x+2x^2+x+1 right)$$

$$beginalign
& C=frac23intfrac1sqrtx^2+x+1left( x^2+x+2 right)left( sqrtx^2+x+1 right)^3du \
& quad =frac23intfracx^2+x+1x^2+x+2du \
& quad =-2intfrac1u^2-7du=frac1sqrt7ln left( fracsqrt7+usqrt7-u right)=frac1sqrt7ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right) \
endalign$$

finally,
$$
int cot^-1left( sqrtx^2+x+1 right) , dx=xcot^-1left( sqrtx^2+x+1 right)+ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)-frac12tan ^-1left( sqrt1+x+x^2 right)-fracsqrt74ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right)
$$

share|cite|improve this answer

edited Aug 9 at 3:39

Michael Hardy

204k23186463

answered Aug 6 at 21:26

user579627

686

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i think this is nice
  – Johnny chad
  Aug 6 at 22:11
  

  
  
  
  

add a comment | 

up vote
3
down vote

If $f(t)=operatornamearccott$, then
$$
f'(t)=-frac11+t^2
$$
so after integration by parts you get
$$
xoperatornamearccotsqrtx^2+x+1+
int xfrac1x^2+x+2frac2x+12sqrtx^2+x+1,dx
$$
For the remaining integral, consider the curve $y=sqrtx^2+x+1$, that's a branch of $x^2-y^2+x+1=0$. This is a hyperbola, which can be put in normal form by completing the square:
$$
left(x+frac12right)^2-y^2+frac34=0
$$
or
$$
-frac(x+1/2)^2(sqrt3/2)^2+fracy^2(sqrt3/2)^2=1
$$
The correct substitution is thus
$$
y=fracsqrt32cosh u
qquad
x=fracsqrt32sinh u-frac12
$$
that will bring the function into a rational function of $e^u$, which is essentially elementary: set $e^u=v$ and use partial fractions.

share|cite|improve this answer

answered Aug 6 at 13:50

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the amazing method. I dont really have a grasp on hyperbolic functions as i have not studied them yet. An alternate solution would be really appreciated( I would like to ask if an alternate solution would even exist or not?)
  – Sidharth Giri
  Aug 6 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SidharthGiri Sorry, I wouldn't really try any other method.
  – egreg
  Aug 6 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, Thanks a lot Sir!
  – Sidharth Giri
  Aug 6 at 17:22
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

A solution without using hyperbolic functions

Integration by parts gives
$$xcot^-1left( sqrtx^2+x+1 right)+frac12int fracx(2x+1)sqrtx^2+x+1(x^2+x+2),dx$$
i can split the integrated on the right as
$$
beginalign
& =2intfrac1sqrtx^2+x+1dx-intfrac1+2x2sqrtx^2+x+1left( x^2+x+2 right)dx-frac72intfrac1sqrt1+x+x^2left( x^2+x+2 right)dx \
& =2A-B-frac72C \
endalign
$$
to calculate $A$ notice that $x^2+x+1=left( x+frac12 right)^2+left( fracsqrt32 right)^2$ and using $x+frac12=fracsqrt32u$

$$A=intfrac1sqrtu^2+1du=ln left( u+sqrtu^2+1 right)=ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)$$
that’s because i accept the fact that $left( ln left( u+sqrtu^2+1 right) right)^prime =frac1sqrtu^2+1$

for $B$ we have
$$frac1+2x2sqrt1+x+x^2left( 2+x+x^2 right)=fracleft( sqrt1+x+x^2 right)^prime left( 1+left( sqrt1+x+x^2 right)^2 right)Rightarrow B=tan ^-1left( sqrt1+x+x^2 right)$$

for $C$ use $u=frac2x+1sqrtx^2+x+1$ hence $du=frac32left( sqrtx^2+x+1 right)^3dx$ also
$$u^2-7=frac4x^2+4x+1x^2+x+1-frac7x^2+7x+7x^2+x+1=frac-3x^2-3x-6x^2+x+1=-3left( fracx^2+x+2x^2+x+1 right)$$

$$beginalign
& C=frac23intfrac1sqrtx^2+x+1left( x^2+x+2 right)left( sqrtx^2+x+1 right)^3du \
& quad =frac23intfracx^2+x+1x^2+x+2du \
& quad =-2intfrac1u^2-7du=frac1sqrt7ln left( fracsqrt7+usqrt7-u right)=frac1sqrt7ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right) \
endalign$$

finally,
$$
int cot^-1left( sqrtx^2+x+1 right) , dx=xcot^-1left( sqrtx^2+x+1 right)+ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)-frac12tan ^-1left( sqrt1+x+x^2 right)-fracsqrt74ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right)
$$

share|cite|improve this answer

edited Aug 9 at 3:39

Michael Hardy

204k23186463

answered Aug 6 at 21:26

user579627

686

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i think this is nice
  – Johnny chad
  Aug 6 at 22:11
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

A solution without using hyperbolic functions

Integration by parts gives
$$xcot^-1left( sqrtx^2+x+1 right)+frac12int fracx(2x+1)sqrtx^2+x+1(x^2+x+2),dx$$
i can split the integrated on the right as
$$
beginalign
& =2intfrac1sqrtx^2+x+1dx-intfrac1+2x2sqrtx^2+x+1left( x^2+x+2 right)dx-frac72intfrac1sqrt1+x+x^2left( x^2+x+2 right)dx \
& =2A-B-frac72C \
endalign
$$
to calculate $A$ notice that $x^2+x+1=left( x+frac12 right)^2+left( fracsqrt32 right)^2$ and using $x+frac12=fracsqrt32u$

$$A=intfrac1sqrtu^2+1du=ln left( u+sqrtu^2+1 right)=ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)$$
that’s because i accept the fact that $left( ln left( u+sqrtu^2+1 right) right)^prime =frac1sqrtu^2+1$

for $B$ we have
$$frac1+2x2sqrt1+x+x^2left( 2+x+x^2 right)=fracleft( sqrt1+x+x^2 right)^prime left( 1+left( sqrt1+x+x^2 right)^2 right)Rightarrow B=tan ^-1left( sqrt1+x+x^2 right)$$

for $C$ use $u=frac2x+1sqrtx^2+x+1$ hence $du=frac32left( sqrtx^2+x+1 right)^3dx$ also
$$u^2-7=frac4x^2+4x+1x^2+x+1-frac7x^2+7x+7x^2+x+1=frac-3x^2-3x-6x^2+x+1=-3left( fracx^2+x+2x^2+x+1 right)$$

$$beginalign
& C=frac23intfrac1sqrtx^2+x+1left( x^2+x+2 right)left( sqrtx^2+x+1 right)^3du \
& quad =frac23intfracx^2+x+1x^2+x+2du \
& quad =-2intfrac1u^2-7du=frac1sqrt7ln left( fracsqrt7+usqrt7-u right)=frac1sqrt7ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right) \
endalign$$

finally,
$$
int cot^-1left( sqrtx^2+x+1 right) , dx=xcot^-1left( sqrtx^2+x+1 right)+ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)-frac12tan ^-1left( sqrt1+x+x^2 right)-fracsqrt74ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right)
$$

share|cite|improve this answer

edited Aug 9 at 3:39

Michael Hardy

204k23186463

answered Aug 6 at 21:26

user579627

686

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i think this is nice
  – Johnny chad
  Aug 6 at 22:11
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

A solution without using hyperbolic functions

Integration by parts gives
$$xcot^-1left( sqrtx^2+x+1 right)+frac12int fracx(2x+1)sqrtx^2+x+1(x^2+x+2),dx$$
i can split the integrated on the right as
$$
beginalign
& =2intfrac1sqrtx^2+x+1dx-intfrac1+2x2sqrtx^2+x+1left( x^2+x+2 right)dx-frac72intfrac1sqrt1+x+x^2left( x^2+x+2 right)dx \
& =2A-B-frac72C \
endalign
$$
to calculate $A$ notice that $x^2+x+1=left( x+frac12 right)^2+left( fracsqrt32 right)^2$ and using $x+frac12=fracsqrt32u$

$$A=intfrac1sqrtu^2+1du=ln left( u+sqrtu^2+1 right)=ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)$$
that’s because i accept the fact that $left( ln left( u+sqrtu^2+1 right) right)^prime =frac1sqrtu^2+1$

for $B$ we have
$$frac1+2x2sqrt1+x+x^2left( 2+x+x^2 right)=fracleft( sqrt1+x+x^2 right)^prime left( 1+left( sqrt1+x+x^2 right)^2 right)Rightarrow B=tan ^-1left( sqrt1+x+x^2 right)$$

for $C$ use $u=frac2x+1sqrtx^2+x+1$ hence $du=frac32left( sqrtx^2+x+1 right)^3dx$ also
$$u^2-7=frac4x^2+4x+1x^2+x+1-frac7x^2+7x+7x^2+x+1=frac-3x^2-3x-6x^2+x+1=-3left( fracx^2+x+2x^2+x+1 right)$$

$$beginalign
& C=frac23intfrac1sqrtx^2+x+1left( x^2+x+2 right)left( sqrtx^2+x+1 right)^3du \
& quad =frac23intfracx^2+x+1x^2+x+2du \
& quad =-2intfrac1u^2-7du=frac1sqrt7ln left( fracsqrt7+usqrt7-u right)=frac1sqrt7ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right) \
endalign$$

finally,
$$
int cot^-1left( sqrtx^2+x+1 right) , dx=xcot^-1left( sqrtx^2+x+1 right)+ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)-frac12tan ^-1left( sqrt1+x+x^2 right)-fracsqrt74ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right)
$$

share|cite|improve this answer

edited Aug 9 at 3:39

Michael Hardy

204k23186463

answered Aug 6 at 21:26

user579627

686

A solution without using hyperbolic functions

Integration by parts gives
$$xcot^-1left( sqrtx^2+x+1 right)+frac12int fracx(2x+1)sqrtx^2+x+1(x^2+x+2),dx$$
i can split the integrated on the right as
$$
beginalign
& =2intfrac1sqrtx^2+x+1dx-intfrac1+2x2sqrtx^2+x+1left( x^2+x+2 right)dx-frac72intfrac1sqrt1+x+x^2left( x^2+x+2 right)dx \
& =2A-B-frac72C \
endalign
$$
to calculate $A$ notice that $x^2+x+1=left( x+frac12 right)^2+left( fracsqrt32 right)^2$ and using $x+frac12=fracsqrt32u$

$$A=intfrac1sqrtu^2+1du=ln left( u+sqrtu^2+1 right)=ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)$$
that’s because i accept the fact that $left( ln left( u+sqrtu^2+1 right) right)^prime =frac1sqrtu^2+1$

for $B$ we have
$$frac1+2x2sqrt1+x+x^2left( 2+x+x^2 right)=fracleft( sqrt1+x+x^2 right)^prime left( 1+left( sqrt1+x+x^2 right)^2 right)Rightarrow B=tan ^-1left( sqrt1+x+x^2 right)$$

for $C$ use $u=frac2x+1sqrtx^2+x+1$ hence $du=frac32left( sqrtx^2+x+1 right)^3dx$ also
$$u^2-7=frac4x^2+4x+1x^2+x+1-frac7x^2+7x+7x^2+x+1=frac-3x^2-3x-6x^2+x+1=-3left( fracx^2+x+2x^2+x+1 right)$$

$$beginalign
& C=frac23intfrac1sqrtx^2+x+1left( x^2+x+2 right)left( sqrtx^2+x+1 right)^3du \
& quad =frac23intfracx^2+x+1x^2+x+2du \
& quad =-2intfrac1u^2-7du=frac1sqrt7ln left( fracsqrt7+usqrt7-u right)=frac1sqrt7ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right) \
endalign$$

finally,
$$
int cot^-1left( sqrtx^2+x+1 right) , dx=xcot^-1left( sqrtx^2+x+1 right)+ln left( frac2x+1sqrt3+sqrtleft( frac2x+1sqrt3 right)^2+1 right)-frac12tan ^-1left( sqrt1+x+x^2 right)-fracsqrt74ln left( fracsqrt7+frac2x+1sqrtx^2+x+1sqrt7-frac2x+1sqrtx^2+x+1 right)
$$

share|cite|improve this answer

edited Aug 9 at 3:39

Michael Hardy

204k23186463

answered Aug 6 at 21:26

user579627

686

share|cite|improve this answer

share|cite|improve this answer

edited Aug 9 at 3:39

Michael Hardy

204k23186463

edited Aug 9 at 3:39

Michael Hardy

204k23186463

edited Aug 9 at 3:39

Michael Hardy

204k23186463

204k23186463

answered Aug 6 at 21:26

user579627

686

answered Aug 6 at 21:26

user579627

686

answered Aug 6 at 21:26

user579627

686

686

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i think this is nice
  – Johnny chad
  Aug 6 at 22:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i think this is nice
  – Johnny chad
  Aug 6 at 22:11
  

  
  
  
  

i think this is nice
– Johnny chad
Aug 6 at 22:11

i think this is nice
– Johnny chad
Aug 6 at 22:11

add a comment | 

up vote
3
down vote

If $f(t)=operatornamearccott$, then
$$
f'(t)=-frac11+t^2
$$
so after integration by parts you get
$$
xoperatornamearccotsqrtx^2+x+1+
int xfrac1x^2+x+2frac2x+12sqrtx^2+x+1,dx
$$
For the remaining integral, consider the curve $y=sqrtx^2+x+1$, that's a branch of $x^2-y^2+x+1=0$. This is a hyperbola, which can be put in normal form by completing the square:
$$
left(x+frac12right)^2-y^2+frac34=0
$$
or
$$
-frac(x+1/2)^2(sqrt3/2)^2+fracy^2(sqrt3/2)^2=1
$$
The correct substitution is thus
$$
y=fracsqrt32cosh u
qquad
x=fracsqrt32sinh u-frac12
$$
that will bring the function into a rational function of $e^u$, which is essentially elementary: set $e^u=v$ and use partial fractions.

share|cite|improve this answer

answered Aug 6 at 13:50

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the amazing method. I dont really have a grasp on hyperbolic functions as i have not studied them yet. An alternate solution would be really appreciated( I would like to ask if an alternate solution would even exist or not?)
  – Sidharth Giri
  Aug 6 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SidharthGiri Sorry, I wouldn't really try any other method.
  – egreg
  Aug 6 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, Thanks a lot Sir!
  – Sidharth Giri
  Aug 6 at 17:22
  

  
  
  
  

add a comment | 

up vote
3
down vote

If $f(t)=operatornamearccott$, then
$$
f'(t)=-frac11+t^2
$$
so after integration by parts you get
$$
xoperatornamearccotsqrtx^2+x+1+
int xfrac1x^2+x+2frac2x+12sqrtx^2+x+1,dx
$$
For the remaining integral, consider the curve $y=sqrtx^2+x+1$, that's a branch of $x^2-y^2+x+1=0$. This is a hyperbola, which can be put in normal form by completing the square:
$$
left(x+frac12right)^2-y^2+frac34=0
$$
or
$$
-frac(x+1/2)^2(sqrt3/2)^2+fracy^2(sqrt3/2)^2=1
$$
The correct substitution is thus
$$
y=fracsqrt32cosh u
qquad
x=fracsqrt32sinh u-frac12
$$
that will bring the function into a rational function of $e^u$, which is essentially elementary: set $e^u=v$ and use partial fractions.

share|cite|improve this answer

answered Aug 6 at 13:50

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the amazing method. I dont really have a grasp on hyperbolic functions as i have not studied them yet. An alternate solution would be really appreciated( I would like to ask if an alternate solution would even exist or not?)
  – Sidharth Giri
  Aug 6 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SidharthGiri Sorry, I wouldn't really try any other method.
  – egreg
  Aug 6 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, Thanks a lot Sir!
  – Sidharth Giri
  Aug 6 at 17:22
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

If $f(t)=operatornamearccott$, then
$$
f'(t)=-frac11+t^2
$$
so after integration by parts you get
$$
xoperatornamearccotsqrtx^2+x+1+
int xfrac1x^2+x+2frac2x+12sqrtx^2+x+1,dx
$$
For the remaining integral, consider the curve $y=sqrtx^2+x+1$, that's a branch of $x^2-y^2+x+1=0$. This is a hyperbola, which can be put in normal form by completing the square:
$$
left(x+frac12right)^2-y^2+frac34=0
$$
or
$$
-frac(x+1/2)^2(sqrt3/2)^2+fracy^2(sqrt3/2)^2=1
$$
The correct substitution is thus
$$
y=fracsqrt32cosh u
qquad
x=fracsqrt32sinh u-frac12
$$
that will bring the function into a rational function of $e^u$, which is essentially elementary: set $e^u=v$ and use partial fractions.

share|cite|improve this answer

answered Aug 6 at 13:50

egreg

164k1180187

If $f(t)=operatornamearccott$, then
$$
f'(t)=-frac11+t^2
$$
so after integration by parts you get
$$
xoperatornamearccotsqrtx^2+x+1+
int xfrac1x^2+x+2frac2x+12sqrtx^2+x+1,dx
$$
For the remaining integral, consider the curve $y=sqrtx^2+x+1$, that's a branch of $x^2-y^2+x+1=0$. This is a hyperbola, which can be put in normal form by completing the square:
$$
left(x+frac12right)^2-y^2+frac34=0
$$
or
$$
-frac(x+1/2)^2(sqrt3/2)^2+fracy^2(sqrt3/2)^2=1
$$
The correct substitution is thus
$$
y=fracsqrt32cosh u
qquad
x=fracsqrt32sinh u-frac12
$$
that will bring the function into a rational function of $e^u$, which is essentially elementary: set $e^u=v$ and use partial fractions.

share|cite|improve this answer

answered Aug 6 at 13:50

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 13:50

egreg

164k1180187

answered Aug 6 at 13:50

egreg

164k1180187

answered Aug 6 at 13:50

egreg

164k1180187

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the amazing method. I dont really have a grasp on hyperbolic functions as i have not studied them yet. An alternate solution would be really appreciated( I would like to ask if an alternate solution would even exist or not?)
  – Sidharth Giri
  Aug 6 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SidharthGiri Sorry, I wouldn't really try any other method.
  – egreg
  Aug 6 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, Thanks a lot Sir!
  – Sidharth Giri
  Aug 6 at 17:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the amazing method. I dont really have a grasp on hyperbolic functions as i have not studied them yet. An alternate solution would be really appreciated( I would like to ask if an alternate solution would even exist or not?)
  – Sidharth Giri
  Aug 6 at 17:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SidharthGiri Sorry, I wouldn't really try any other method.
  – egreg
  Aug 6 at 17:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright, Thanks a lot Sir!
  – Sidharth Giri
  Aug 6 at 17:22
  

  
  
  
  

Thanks for the amazing method. I dont really have a grasp on hyperbolic functions as i have not studied them yet. An alternate solution would be really appreciated( I would like to ask if an alternate solution would even exist or not?)
– Sidharth Giri
Aug 6 at 17:19

Thanks for the amazing method. I dont really have a grasp on hyperbolic functions as i have not studied them yet. An alternate solution would be really appreciated( I would like to ask if an alternate solution would even exist or not?)
– Sidharth Giri
Aug 6 at 17:19

@SidharthGiri Sorry, I wouldn't really try any other method.
– egreg
Aug 6 at 17:21

@SidharthGiri Sorry, I wouldn't really try any other method.
– egreg
Aug 6 at 17:21

Alright, Thanks a lot Sir!
– Sidharth Giri
Aug 6 at 17:22

Alright, Thanks a lot Sir!
– Sidharth Giri
Aug 6 at 17:22

add a comment |