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Finding an integral using a table?

Integral of $cosleft(frac1xright), dx$How to solve this indefinite integral using integral substitution?Two solutions for the same integral question, which approach is correct/better to solve?Solving integral without simplifying equationFinding the integral of $x^2sqrt[3]{1-x}$How to find the value of this indefinite integral?Finding double integral of this region using polar coordinates?Solving Integral with Symbolic Computation (Sympy), Division and Tricky LimitsProving Table of Integral Integral (Trigonometric Substitution)Using a table of integrals for solving these integrals

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