Can You Solve This Integral Without Using Integration by Parts?

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Homework Help Overview

The discussion revolves around evaluating the integral of the function x^3/sqrt[x^2 + 1] over the interval [0,1]. The subject area is integral calculus, specifically focusing on techniques for integration.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of integration by parts and the challenges in selecting appropriate u and dv terms. Some suggest rewriting the integral to facilitate this choice, while others propose a substitution method as an alternative approach.

Discussion Status

The conversation includes various strategies for tackling the integral, with some participants exploring integration by parts and others questioning whether this method is necessary. There is no explicit consensus on the preferred method, but multiple approaches are being considered.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to evaluate the integral without relying solely on integration by parts.

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"Evaluate the integral [0,1] x^3/sqrt[x^2 + 1] by integration by parts"

I know I have to use the integration by parts equation, but I don't know what to make u and what to make dv..
 
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Try rewriting the integral thus;

[tex]\int^{1}_{0}\; \frac{x^{3}}{\sqrt{x^2+1}} \; dx = \int^{1}_{0}\; x^3 (x^2 +1)^{-\frac{1}{2}} \;dx[/tex]

Now, to determine which term to make u and dv, think about which one will simplify your expression most when you differentiated it and set this to u.
 
You could start by rewriting the integrand:

[tex] \frac{x^3}{\sqrt{x^2+1}}=x^3 (x^2+1)^{-1/2} = [x^{-6} (x^2+1)]^{-1/2}[/tex]

And the apply the integration by parts formula. That should make it easier.
 
OH okay, thank you!
 
Do you have to do it by parts? Much easier to substitute [tex]u = x^2[/tex]
 

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