Quick integration question

  • Thread starter binbagsss
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  • #1
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Homework Statement



Apologies if this is obvious, maybe I'm a little out of touch

## \int\limits^b_0 \frac{x^3}{x^2+m^2} dx ##

Homework Equations




The Attempt at a Solution



I [/B]was going to go by parts breaking the ##x^3 = x^2 . x##

So that I have the logarithm

I.e :

##b^2 \frac{log (b^2 + m^2)}{2} - \int \frac{log (x^2+m^2)}{2} 2x dx ##



But the solution is :

## b^2 + m^2 log ( \frac{m^2}{b^2+m^2} ) ##


( I thought that perhaps the solution could be going by parts again, but there is no reason for the boundary term to vanish )

Ta
 

Answers and Replies

  • #2
34,688
6,394

Homework Statement



Apologies if this is obvious, maybe I'm a little out of touch

## \int\limits^b_0 \frac{x^3}{x^2+m^2} dx ##

Homework Equations




The Attempt at a Solution



I [/B]was going to go by parts breaking the ##x^3 = x^2 . x##

So that I have the logarithm

I.e :

##b^2 \frac{log (b^2 + m^2)}{2} - \int \frac{log (x^2+m^2)}{2} 2x dx ##



But the solution is :

## b^2 + m^2 log ( \frac{m^2}{b^2+m^2} ) ##


( I thought that perhaps the solution could be going by parts again, but there is no reason for the boundary term to vanish )

Ta
The second term in your answer can be integrated using substitution, with ##u = x^2 + b^2, du = 2xdx##.
 
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  • #3
Delta2
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Write ##x^3=x(x^2+m^2)-m^2x## so the integral becomes ##\int xdx-\int \frac{m^2x}{x^2+m^2}dx##.
 
  • #4
vela
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Or just use the substitution ##u=x^2+m^2## on the original integral.
 

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