Integrate by Parts: Solving \int \ln (x^2 + 1) \, dx

In summary, the conversation discusses finding or evaluating an integral using substitution and integration by parts. The attempt at a solution involves setting u = x^2 + 1 and using integration by parts, but the person is unsure if they are approaching it correctly. They are then given a suggestion to integrate using parts with u=ln(1+x^2) and v=x.
  • #1
GunnaSix
35
0

Homework Statement


Find or evaluate the integral using substitution first, then using integration by parts.

[tex]\int \ln (x^2 + 1) \, dx [/tex]


The Attempt at a Solution



[tex]Let \: u = x^2 + 1[/tex]

[tex]du = 2x \, dx[/tex]

[tex]dx = \pm \frac{du}{2 \sqrt{u - 1}}[/tex]

Then

[tex]\int \ln (x^2 + 1) \, dx = \pm \frac{1}{2} \int \frac{\ln u}{\sqrt{u-1}}\, du[/tex]

I don't know where to go from here. I tried to integrate by parts and it just turned into a mess. Am I approaching this the wrong way?
 
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  • #2
I'm not sure what substitution they think you ought to make. It's pretty easy if you apply parts right off regarding the original integral as u*dv where u=ln(1+x^2) and v=x.
 
  • #3
Got it. Thanks.
 

1. What is integration by parts?

Integration by parts is a mathematical method used to simplify and solve integrals that involve the product of two functions. It involves using the product rule from differentiation to rewrite the integral in a simpler form.

2. How do you use integration by parts to solve \int \ln (x^2+1) \, dx?

To solve this integral, we use the formula: \int u \, dv = uv - \int v \, du. We let u = \ln (x^2+1) and dv = dx, which means du = \frac{2x}{x^2+1} \, dx and v = x. Substituting these values into the formula, we get:\int \ln (x^2+1) \, dx = x \ln (x^2+1) - \int \frac{2x^2}{x^2+1} \, dxWe can then use a trigonometric substitution to solve the remaining integral.

3. When do you use integration by parts?

Integration by parts is typically used when the integral involves the product of two functions, and one of the functions becomes simpler after differentiation. It is also useful when trying to integrate logarithmic, inverse trigonometric, or exponential functions.

4. What are the benefits of using integration by parts?

Integration by parts allows us to solve integrals that may be otherwise difficult or impossible to solve using other methods. It also helps us to simplify integrals and make them easier to solve.

5. Are there any limitations to using integration by parts?

Integration by parts can only be used on integrals that involve the product of two functions. It also requires some knowledge of differentiation and algebraic manipulation, so it may not be the best method for beginners or those who are not familiar with these concepts.

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