When is the integral of $x^-1 = ln lvert x rvert$? [closed]

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In my textbook there is a question that says:

Use either a suitable substitution or the reverse chain rule to find the following integrals.

$$ int left(fracxsqrtx^2 + 2right) dx$$

Using substitution I came down to:

$$1over 2 int u^-1over2du$$
where $u=x^2+2$

Couldn't this be rewritten as $$1over 2 int (sqrtu)^-1$$ And isn't that equivalent to $$1over 2log|sqrtu| + c$$ where $sqrtu$ acts as $x$?

The thing is the answer is $sqrtx^2 +2 + c$. And so I don't know if I fully understand the theory behind integrating $1/x$ and $log|x|$.

Thanks!

integration substitution

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edited Jul 26 at 17:44

Davide Morgante

1,751220

asked Jul 26 at 17:26

Cheks Nweze

627

closed as off-topic by amWhy, hardmath, Isaac Browne, Lord Shark the Unknown, Jyrki Lahtonen Jul 27 at 5:57

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, Lord Shark the Unknown, Jyrki Lahtonen

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Please see math.meta.stackexchange.com/questions/5020 .
  – Lord Shark the Unknown
  Jul 26 at 17:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What is going on? Is LaTeX going out of fashion?
  – mvw
  Jul 26 at 17:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're right that $∫ 1/x dx = ln|x| + C$. How you then got to $∫ 1/sqrtx dx = ln|sqrtx| + C$ is beyond me.
  – Kaynex
  Jul 26 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You've omitted a step in the $u$-substitution, coincident with dropping the $dx,du$ notation from integrals.
  – hardmath
  Jul 27 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it only applies where the power on the x is $-1$ you cannot split the power up like this and treat it as the same function
  – Henry Lee
  Jul 27 at 11:18
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

In my textbook there is a question that says:

Use either a suitable substitution or the reverse chain rule to find the following integrals.

$$ int left(fracxsqrtx^2 + 2right) dx$$

Using substitution I came down to:

$$1over 2 int u^-1over2du$$
where $u=x^2+2$

Couldn't this be rewritten as $$1over 2 int (sqrtu)^-1$$ And isn't that equivalent to $$1over 2log|sqrtu| + c$$ where $sqrtu$ acts as $x$?

The thing is the answer is $sqrtx^2 +2 + c$. And so I don't know if I fully understand the theory behind integrating $1/x$ and $log|x|$.

Thanks!

integration substitution

share|cite|improve this question

edited Jul 26 at 17:44

Davide Morgante

1,751220

asked Jul 26 at 17:26

Cheks Nweze

627

closed as off-topic by amWhy, hardmath, Isaac Browne, Lord Shark the Unknown, Jyrki Lahtonen Jul 27 at 5:57

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, Lord Shark the Unknown, Jyrki Lahtonen

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Please see math.meta.stackexchange.com/questions/5020 .
  – Lord Shark the Unknown
  Jul 26 at 17:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What is going on? Is LaTeX going out of fashion?
  – mvw
  Jul 26 at 17:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're right that $∫ 1/x dx = ln|x| + C$. How you then got to $∫ 1/sqrtx dx = ln|sqrtx| + C$ is beyond me.
  – Kaynex
  Jul 26 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You've omitted a step in the $u$-substitution, coincident with dropping the $dx,du$ notation from integrals.
  – hardmath
  Jul 27 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it only applies where the power on the x is $-1$ you cannot split the power up like this and treat it as the same function
  – Henry Lee
  Jul 27 at 11:18
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

up vote
0
down vote

favorite

In my textbook there is a question that says:

Use either a suitable substitution or the reverse chain rule to find the following integrals.

$$ int left(fracxsqrtx^2 + 2right) dx$$

Using substitution I came down to:

$$1over 2 int u^-1over2du$$
where $u=x^2+2$

Couldn't this be rewritten as $$1over 2 int (sqrtu)^-1$$ And isn't that equivalent to $$1over 2log|sqrtu| + c$$ where $sqrtu$ acts as $x$?

The thing is the answer is $sqrtx^2 +2 + c$. And so I don't know if I fully understand the theory behind integrating $1/x$ and $log|x|$.

Thanks!

integration substitution

share|cite|improve this question

edited Jul 26 at 17:44

Davide Morgante

1,751220

asked Jul 26 at 17:26

Cheks Nweze

627

In my textbook there is a question that says:

Use either a suitable substitution or the reverse chain rule to find the following integrals.

$$ int left(fracxsqrtx^2 + 2right) dx$$

Using substitution I came down to:

$$1over 2 int u^-1over2du$$
where $u=x^2+2$

Couldn't this be rewritten as $$1over 2 int (sqrtu)^-1$$ And isn't that equivalent to $$1over 2log|sqrtu| + c$$ where $sqrtu$ acts as $x$?

The thing is the answer is $sqrtx^2 +2 + c$. And so I don't know if I fully understand the theory behind integrating $1/x$ and $log|x|$.

Thanks!

integration substitution

share|cite|improve this question

edited Jul 26 at 17:44

Davide Morgante

1,751220

asked Jul 26 at 17:26

Cheks Nweze

627

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 17:44

Davide Morgante

1,751220

edited Jul 26 at 17:44

Davide Morgante

1,751220

edited Jul 26 at 17:44

Davide Morgante

1,751220

1,751220

asked Jul 26 at 17:26

Cheks Nweze

627

asked Jul 26 at 17:26

Cheks Nweze

627

asked Jul 26 at 17:26

Cheks Nweze

627

627

closed as off-topic by amWhy, hardmath, Isaac Browne, Lord Shark the Unknown, Jyrki Lahtonen Jul 27 at 5:57

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, Lord Shark the Unknown, Jyrki Lahtonen

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

closed as off-topic by amWhy, hardmath, Isaac Browne, Lord Shark the Unknown, Jyrki Lahtonen Jul 27 at 5:57

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, Lord Shark the Unknown, Jyrki Lahtonen

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

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Please see math.meta.stackexchange.com/questions/5020 .
  – Lord Shark the Unknown
  Jul 26 at 17:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What is going on? Is LaTeX going out of fashion?
  – mvw
  Jul 26 at 17:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're right that $∫ 1/x dx = ln|x| + C$. How you then got to $∫ 1/sqrtx dx = ln|sqrtx| + C$ is beyond me.
  – Kaynex
  Jul 26 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You've omitted a step in the $u$-substitution, coincident with dropping the $dx,du$ notation from integrals.
  – hardmath
  Jul 27 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it only applies where the power on the x is $-1$ you cannot split the power up like this and treat it as the same function
  – Henry Lee
  Jul 27 at 11:18
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Please see math.meta.stackexchange.com/questions/5020 .
  – Lord Shark the Unknown
  Jul 26 at 17:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What is going on? Is LaTeX going out of fashion?
  – mvw
  Jul 26 at 17:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're right that $∫ 1/x dx = ln|x| + C$. How you then got to $∫ 1/sqrtx dx = ln|sqrtx| + C$ is beyond me.
  – Kaynex
  Jul 26 at 17:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You've omitted a step in the $u$-substitution, coincident with dropping the $dx,du$ notation from integrals.
  – hardmath
  Jul 27 at 2:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it only applies where the power on the x is $-1$ you cannot split the power up like this and treat it as the same function
  – Henry Lee
  Jul 27 at 11:18
  

  
  
  
  

2

2

Please see math.meta.stackexchange.com/questions/5020 .
– Lord Shark the Unknown
Jul 26 at 17:26

Please see math.meta.stackexchange.com/questions/5020 .
– Lord Shark the Unknown
Jul 26 at 17:26

1

1

What is going on? Is LaTeX going out of fashion?
– mvw
Jul 26 at 17:35

What is going on? Is LaTeX going out of fashion?
– mvw
Jul 26 at 17:35

You're right that $∫ 1/x dx = ln|x| + C$. How you then got to $∫ 1/sqrtx dx = ln|sqrtx| + C$ is beyond me.
– Kaynex
Jul 26 at 17:41

You're right that $∫ 1/x dx = ln|x| + C$. How you then got to $∫ 1/sqrtx dx = ln|sqrtx| + C$ is beyond me.
– Kaynex
Jul 26 at 17:41

You've omitted a step in the $u$-substitution, coincident with dropping the $dx,du$ notation from integrals.
– hardmath
Jul 27 at 2:03

You've omitted a step in the $u$-substitution, coincident with dropping the $dx,du$ notation from integrals.
– hardmath
Jul 27 at 2:03

it only applies where the power on the x is $-1$ you cannot split the power up like this and treat it as the same function
– Henry Lee
Jul 27 at 11:18

it only applies where the power on the x is $-1$ you cannot split the power up like this and treat it as the same function
– Henry Lee
Jul 27 at 11:18

 | 
show 1 more comment

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

$int frac 1u du = ln |u|+C.$ However, $int frac 1f(u) du ne ln |f(u)|$

That is, you can only integate into a logarthm if, by u-substitutions you can get the denominator with that lone $u.$

For this problem we have a power rule you can employ:

$int u^a du = frac u^a+1a+1 + C$ (which works for all constants $ane -1$ )

share|cite|improve this answer

answered Jul 26 at 17:39

Doug M

39k31749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, right. But if I was to integrate $int_^frac12u+4du$ wouldn't I get $frac12ln|2u+4|+c$? Or is it because I can rewrite $frac12u+4$ as $frac1u$ plus something else?
  – Cheks Nweze
  Jul 26 at 18:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int frac 1ax+b dx = frac 1a ln |ax+b| +C$ but that is because you are able to do the substitution $u = ax+b$ to isolate $u$ in the denominator. I would say stick with this extra step until you are more comfortable knowing when you can combine the two steps at once.
  – Doug M
  Jul 26 at 18:04
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

We have that

• for $x>0$ $$f(x)=ln |x|=ln ximplies f’(x)=frac1x$$




• for $x<0$ $$g(x)=ln |x|=ln (-x)implies g’(x)=-frac1-x=frac1x$$

share|cite|improve this answer

answered Jul 26 at 17:37

gimusi

65k73583

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

$int frac 1u du = ln |u|+C.$ However, $int frac 1f(u) du ne ln |f(u)|$

That is, you can only integate into a logarthm if, by u-substitutions you can get the denominator with that lone $u.$

For this problem we have a power rule you can employ:

$int u^a du = frac u^a+1a+1 + C$ (which works for all constants $ane -1$ )

share|cite|improve this answer

answered Jul 26 at 17:39

Doug M

39k31749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, right. But if I was to integrate $int_^frac12u+4du$ wouldn't I get $frac12ln|2u+4|+c$? Or is it because I can rewrite $frac12u+4$ as $frac1u$ plus something else?
  – Cheks Nweze
  Jul 26 at 18:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int frac 1ax+b dx = frac 1a ln |ax+b| +C$ but that is because you are able to do the substitution $u = ax+b$ to isolate $u$ in the denominator. I would say stick with this extra step until you are more comfortable knowing when you can combine the two steps at once.
  – Doug M
  Jul 26 at 18:04
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

$int frac 1u du = ln |u|+C.$ However, $int frac 1f(u) du ne ln |f(u)|$

That is, you can only integate into a logarthm if, by u-substitutions you can get the denominator with that lone $u.$

For this problem we have a power rule you can employ:

$int u^a du = frac u^a+1a+1 + C$ (which works for all constants $ane -1$ )

share|cite|improve this answer

answered Jul 26 at 17:39

Doug M

39k31749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, right. But if I was to integrate $int_^frac12u+4du$ wouldn't I get $frac12ln|2u+4|+c$? Or is it because I can rewrite $frac12u+4$ as $frac1u$ plus something else?
  – Cheks Nweze
  Jul 26 at 18:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int frac 1ax+b dx = frac 1a ln |ax+b| +C$ but that is because you are able to do the substitution $u = ax+b$ to isolate $u$ in the denominator. I would say stick with this extra step until you are more comfortable knowing when you can combine the two steps at once.
  – Doug M
  Jul 26 at 18:04
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

$int frac 1u du = ln |u|+C.$ However, $int frac 1f(u) du ne ln |f(u)|$

That is, you can only integate into a logarthm if, by u-substitutions you can get the denominator with that lone $u.$

For this problem we have a power rule you can employ:

$int u^a du = frac u^a+1a+1 + C$ (which works for all constants $ane -1$ )

share|cite|improve this answer

answered Jul 26 at 17:39

Doug M

39k31749

$int frac 1u du = ln |u|+C.$ However, $int frac 1f(u) du ne ln |f(u)|$

That is, you can only integate into a logarthm if, by u-substitutions you can get the denominator with that lone $u.$

For this problem we have a power rule you can employ:

$int u^a du = frac u^a+1a+1 + C$ (which works for all constants $ane -1$ )

share|cite|improve this answer

answered Jul 26 at 17:39

Doug M

39k31749

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 17:39

Doug M

39k31749

answered Jul 26 at 17:39

Doug M

39k31749

answered Jul 26 at 17:39

Doug M

39k31749

39k31749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, right. But if I was to integrate $int_^frac12u+4du$ wouldn't I get $frac12ln|2u+4|+c$? Or is it because I can rewrite $frac12u+4$ as $frac1u$ plus something else?
  – Cheks Nweze
  Jul 26 at 18:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int frac 1ax+b dx = frac 1a ln |ax+b| +C$ but that is because you are able to do the substitution $u = ax+b$ to isolate $u$ in the denominator. I would say stick with this extra step until you are more comfortable knowing when you can combine the two steps at once.
  – Doug M
  Jul 26 at 18:04
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, right. But if I was to integrate $int_^frac12u+4du$ wouldn't I get $frac12ln|2u+4|+c$? Or is it because I can rewrite $frac12u+4$ as $frac1u$ plus something else?
  – Cheks Nweze
  Jul 26 at 18:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int frac 1ax+b dx = frac 1a ln |ax+b| +C$ but that is because you are able to do the substitution $u = ax+b$ to isolate $u$ in the denominator. I would say stick with this extra step until you are more comfortable knowing when you can combine the two steps at once.
  – Doug M
  Jul 26 at 18:04
  
  

  
  
  
  

Ah, right. But if I was to integrate $int_^frac12u+4du$ wouldn't I get $frac12ln|2u+4|+c$? Or is it because I can rewrite $frac12u+4$ as $frac1u$ plus something else?
– Cheks Nweze
Jul 26 at 18:00

Ah, right. But if I was to integrate $int_^frac12u+4du$ wouldn't I get $frac12ln|2u+4|+c$? Or is it because I can rewrite $frac12u+4$ as $frac1u$ plus something else?
– Cheks Nweze
Jul 26 at 18:00

$int frac 1ax+b dx = frac 1a ln |ax+b| +C$ but that is because you are able to do the substitution $u = ax+b$ to isolate $u$ in the denominator. I would say stick with this extra step until you are more comfortable knowing when you can combine the two steps at once.
– Doug M
Jul 26 at 18:04

$int frac 1ax+b dx = frac 1a ln |ax+b| +C$ but that is because you are able to do the substitution $u = ax+b$ to isolate $u$ in the denominator. I would say stick with this extra step until you are more comfortable knowing when you can combine the two steps at once.
– Doug M
Jul 26 at 18:04

add a comment | 

up vote
0
down vote

We have that

• for $x>0$ $$f(x)=ln |x|=ln ximplies f’(x)=frac1x$$




• for $x<0$ $$g(x)=ln |x|=ln (-x)implies g’(x)=-frac1-x=frac1x$$

share|cite|improve this answer

answered Jul 26 at 17:37

gimusi

65k73583

add a comment | 

up vote
0
down vote

We have that

• for $x>0$ $$f(x)=ln |x|=ln ximplies f’(x)=frac1x$$




• for $x<0$ $$g(x)=ln |x|=ln (-x)implies g’(x)=-frac1-x=frac1x$$

share|cite|improve this answer

answered Jul 26 at 17:37

gimusi

65k73583

add a comment | 

up vote
0
down vote

up vote
0
down vote

We have that

• for $x>0$ $$f(x)=ln |x|=ln ximplies f’(x)=frac1x$$




• for $x<0$ $$g(x)=ln |x|=ln (-x)implies g’(x)=-frac1-x=frac1x$$

share|cite|improve this answer

answered Jul 26 at 17:37

gimusi

65k73583

We have that

• for $x>0$ $$f(x)=ln |x|=ln ximplies f’(x)=frac1x$$




• for $x<0$ $$g(x)=ln |x|=ln (-x)implies g’(x)=-frac1-x=frac1x$$

share|cite|improve this answer

answered Jul 26 at 17:37

gimusi

65k73583

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 17:37

gimusi

65k73583

answered Jul 26 at 17:37

gimusi

65k73583

answered Jul 26 at 17:37

gimusi

65k73583

65k73583

add a comment | 

add a comment |