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

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SUMMARY

The integral \(\int \ln (x^2 + 1) \, dx\) can be effectively solved using integration by parts. The correct approach involves letting \(u = \ln(x^2 + 1)\) and \(dv = dx\), leading to \(du = \frac{2x}{x^2 + 1} \, dx\) and \(v = x\). This method simplifies the integration process and avoids complications that arise from incorrect substitutions. The discussion highlights the importance of choosing the right \(u\) and \(dv\) in integration by parts.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with logarithmic functions and their properties.
  • Basic knowledge of substitution methods in calculus.
  • Ability to manipulate algebraic expressions involving derivatives.
NEXT STEPS
  • Practice solving integrals using integration by parts with different functions.
  • Explore advanced substitution techniques in integral calculus.
  • Review the properties of logarithmic functions to enhance integration skills.
  • Study common integral forms and their solutions for quicker problem-solving.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for effective teaching methods for integration by parts.

GunnaSix
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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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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.
 
Got it. Thanks.
 

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