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Finding an integral using a table?

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2

2

$begingroup$

Am I correct for pattern matching this integral?

I have

$$int frac{sqrt{9x^2+4}}{x^2}dx$$

Does this pattern match with:

$$int frac{sqrt{a^2 + x^2}}{x^2}dx = -frac{a^2 + x^2}{x} + ln(x + sqrt{a^2 + x^2}) + c$$

If I factor out the 9, I get

$$= 3 int frac{sqrt{x^2 + frac{4}{9}}}{x^2}$$
with $a = frac{2}{3}$

I get:
$$3 left( - frac{sqrt{frac{4}{9}+x^2}}{x} + lnleft(x+sqrt{frac{4}{9}+x^2}right) +cright)$$

Is this the right track?

Wolfram winds up with a different answer though:

integration

share|cite|improve this question

edited 1 hour ago

clathratus

4,745337

asked 3 hours ago

Jwan622Jwan622

2,20611632

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried manipulating your answer to look like Wolfram's? Also, note that it's fine if your answer differs from Wolfram's by a constant.
  $endgroup$
  – Minus One-Twelfth
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
  $endgroup$
  – xbh
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  One quick way to check the result is to put your formula into Wolfram Alpha (without the constant, which in this case is just adding $3c$ to the result) and subtract the formula Wolfram Alpha gave (again omitting the constant). If the result is a flat constant function then your integral is correct.
  $endgroup$
  – David K
  1 hour ago
  

  
  
  
  

add a comment | 

2

2

$begingroup$

Am I correct for pattern matching this integral?

I have

$$int frac{sqrt{9x^2+4}}{x^2}dx$$

Does this pattern match with:

$$int frac{sqrt{a^2 + x^2}}{x^2}dx = -frac{a^2 + x^2}{x} + ln(x + sqrt{a^2 + x^2}) + c$$

If I factor out the 9, I get

$$= 3 int frac{sqrt{x^2 + frac{4}{9}}}{x^2}$$
with $a = frac{2}{3}$

I get:
$$3 left( - frac{sqrt{frac{4}{9}+x^2}}{x} + lnleft(x+sqrt{frac{4}{9}+x^2}right) +cright)$$

Is this the right track?

Wolfram winds up with a different answer though:

integration

share|cite|improve this question

edited 1 hour ago

clathratus

4,745337

asked 3 hours ago

Jwan622Jwan622

2,20611632

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried manipulating your answer to look like Wolfram's? Also, note that it's fine if your answer differs from Wolfram's by a constant.
  $endgroup$
  – Minus One-Twelfth
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
  $endgroup$
  – xbh
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  One quick way to check the result is to put your formula into Wolfram Alpha (without the constant, which in this case is just adding $3c$ to the result) and subtract the formula Wolfram Alpha gave (again omitting the constant). If the result is a flat constant function then your integral is correct.
  $endgroup$
  – David K
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Am I correct for pattern matching this integral?

I have

$$int frac{sqrt{9x^2+4}}{x^2}dx$$

Does this pattern match with:

$$int frac{sqrt{a^2 + x^2}}{x^2}dx = -frac{a^2 + x^2}{x} + ln(x + sqrt{a^2 + x^2}) + c$$

If I factor out the 9, I get

$$= 3 int frac{sqrt{x^2 + frac{4}{9}}}{x^2}$$
with $a = frac{2}{3}$

I get:
$$3 left( - frac{sqrt{frac{4}{9}+x^2}}{x} + lnleft(x+sqrt{frac{4}{9}+x^2}right) +cright)$$

Is this the right track?

Wolfram winds up with a different answer though:

integration

share|cite|improve this question

edited 1 hour ago

clathratus

4,745337

asked 3 hours ago

Jwan622Jwan622

2,20611632

$endgroup$

share|cite|improve this question

edited 1 hour ago

clathratus

4,745337

asked 3 hours ago

Jwan622Jwan622

2,20611632

share|cite|improve this question

edited 1 hour ago

clathratus

4,745337

asked 3 hours ago

Jwan622Jwan622

2,20611632

edited 1 hour ago

clathratus

4,745337

edited 1 hour ago

clathratus

4,745337

asked 3 hours ago

Jwan622Jwan622

2,20611632

asked 3 hours ago

Jwan622Jwan622

2,20611632

1 Answer
1

active

oldest

votes

4

$begingroup$

You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.

The first term in Wolfram's answer can be rewritten:

$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$

and the second term can be rearranged to be identical to your other term.

So your answers are separated by a constant. That's fine. You're right.

share|cite|improve this answer

answered 3 hours ago

DeepakDeepak

17.3k11537

$endgroup$

add a comment | 

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4

$begingroup$

You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.

The first term in Wolfram's answer can be rewritten:

$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$

and the second term can be rearranged to be identical to your other term.

So your answers are separated by a constant. That's fine. You're right.

share|cite|improve this answer

answered 3 hours ago

DeepakDeepak

17.3k11537

$endgroup$

add a comment | 

4

$begingroup$

You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.

The first term in Wolfram's answer can be rewritten:

$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$

and the second term can be rearranged to be identical to your other term.

So your answers are separated by a constant. That's fine. You're right.

share|cite|improve this answer

answered 3 hours ago

DeepakDeepak

17.3k11537

$endgroup$

add a comment | 

$begingroup$

You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.

The first term in Wolfram's answer can be rewritten:

$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$

and the second term can be rearranged to be identical to your other term.

So your answers are separated by a constant. That's fine. You're right.

share|cite|improve this answer

answered 3 hours ago

DeepakDeepak

17.3k11537

$endgroup$

share|cite|improve this answer

answered 3 hours ago

DeepakDeepak

17.3k11537

answered 3 hours ago

DeepakDeepak

17.3k11537

answered 3 hours ago

DeepakDeepak

17.3k11537