Integration by Parts for ln(sqrt(x^2-4)) | Solving Tricky Integration Problems

In summary, to integrate ln(sqrt(x^2-4)), u-substitution will not work and integration by parts is the suggested method. The attempt at a solution involves using u = 1/2ln(x^2-4) and dv = dx, leading to a solution involving partial fractions and log properties.
  • #1
Bryon
99
0

Homework Statement



The the problem is integrate ln(sqrt(x^2-4)

Homework Equations



For this one u-sub does not work, which leads me to think that integration by parts is the way to go but am not sure how to finish it.



The Attempt at a Solution



u =1/2ln(x^2-4)
du = (1/2)(2x/(x^2-4)) = x^2/(x^2-4)

dv = dx ----> int dx = x = v

x(1/2ln(x^2-4)) - int (x^2)/(X^2-4)

I don't know what to do about the part in bold. I am thinking integration by part again but that doesn't look like it will work, unless there is one more integration to do. Any idea what i can try?

Thanks!
 
Physics news on Phys.org
  • #2
I think parts can work again, since 1 / x^2 - 4 is easy enough to integrate, then you'll need to integrate xarctan(x/2) which should be fine
 
  • #3
Use polynomial division to write x^2/(x^2-4)=1+4/(x^2-4). Looks more to me like 'partial fractions' than 'parts'.
 
  • #4
Write it as
ln(x^2 - 4)^1/2

and also note that it's a difference of 2 squares. Then just use the log properties to reduce this to something easier and you can then use integration by parts.
 
  • #5
Thanks Dave! I did not see that, but it certainlly makes sense now.
 

What is integration by parts?

Integration by parts is a mathematical technique used to evaluate integrals involving the product of two functions. It is based on the product rule of differentiation and allows us to simplify complex integrals into smaller, more manageable ones.

When should I use integration by parts?

You should use integration by parts when you have an integral that involves the product of two functions and one of the functions can be easily integrated while the other can be easily differentiated. This technique is especially useful for solving integrals involving logarithmic, exponential, and trigonometric functions.

What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are the two functions being multiplied together and du and dv are their respective differentials.

What is the process for integration by parts?

The process for integration by parts involves choosing u and dv, computing du and v, plugging them into the formula, and then simplifying the resulting integral. This process may need to be repeated multiple times until the integral can be solved.

Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for definite integrals. After finding the antiderivative using the integration by parts formula, the limits of integration can be substituted into the resulting equation to find the definite integral.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
419
  • Calculus and Beyond Homework Help
Replies
15
Views
776
  • Calculus and Beyond Homework Help
Replies
8
Views
754
  • Calculus and Beyond Homework Help
Replies
3
Views
571
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
729
  • Calculus and Beyond Homework Help
Replies
7
Views
692
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Back
Top