Can I Solve x^4 dx / (x^2+a^2)^(3/2)?

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Homework Help Overview

The discussion revolves around the integral of the function x^4 dx / (x^2 + a^2)^(3/2), exploring various methods of integration and substitutions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different substitution methods, including trigonometric substitutions and hyperbolic functions. Some express difficulties in simplifying the integral after substitution, while others suggest alternative approaches.

Discussion Status

There is an ongoing exploration of various substitution techniques, with participants sharing their progress and challenges. Some guidance has been offered regarding potential substitutions, but no consensus or resolution has been reached.

Contextual Notes

Participants mention specific limits of integration and transformations related to the substitutions they are attempting. There is a recognition of the complexity of the integral and the challenges posed by the various approaches discussed.

theneedtoknow
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Is this integral doable at all ?

x^4 dx / (x^2+a^2)^(3/2)

I have tried several u-substitutions but none of them seem to make the form of the integral any simpler
 
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I looks doable to me. Did you try a trig substitution like x=a*tan(t)?
 
Thanks :)
so this is how far I get:

if x=atan(u)
then dx=asec^2(u)du
my numerator becomes x^4dx = a^5tan^4(u)sec^2(u)du

since (x^2+a^2)^1/2 = asec(u)
then my denominator will be a^3sec^3(u)

my old limits of integration were x=0 to x=R
new ones become u=0 to u=tan^-1(R/a)

so my new integral is:

integral from 0 to tan^-1(R/a) of a^2tan^4(u)cos(u)du

unfortunately this is where I get suck again. :S I tried some kind of tan^2=sec^2-1 substitution but I couldn't figure out how to do the resulting integral either
 
theneedtoknow said:
Thanks :)
so this is how far I get:

if x=atan(u)
then dx=asec^2(u)du
my numerator becomes x^4dx = a^5tan^4(u)sec^2(u)du

since (x^2+a^2)^1/2 = asec(u)
then my denominator will be a^3sec^3(u)

my old limits of integration were x=0 to x=R
new ones become u=0 to u=tan^-1(R/a)

so my new integral is:

integral from 0 to tan^-1(R/a) of a^2tan^4(u)cos(u)du

unfortunately this is where I get suck again. :S I tried some kind of tan^2=sec^2-1 substitution but I couldn't figure out how to do the resulting integral either

Now try v=sin(u) and remember tan(u)=sin(u)/cos(u). I said it was doable. I didn't say it was easy.
 
So , if I let v=sinu, then dv=cosudu ---> du=dv/cosu >> tan^4ucosu = v^4dv /cos^4(u) = v^4 dv /(1-v^2)^2

Do i now use long division to reduce the power of the denominator by one, and then solve the remainder with partial fractions?
 
theneedtoknow said:
So , if I let v=sinu, then dv=cosudu ---> du=dv/cosu >> tan^4ucosu = v^4dv /cos^4(u) = v^4 dv /(1-v^2)^2

Do i now use long division to reduce the power of the denominator by one, and then solve the remainder with partial fractions?

Right.
 
Wow thank you so much!

That was one hell of an integral..!
 
x = a sinh(t) is a lot simpler. With some effort you can see the answer using mental computations, without using paper and pencil.
 

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