How Can You Integrate x/(a^2+x^2)^(3/2) Without Explicit Substitution?

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In summary, in the video, the professor integrated x/(a^2+x^2)^3/2 by ignoring the x and treating the integrand as 1/(a^2+x^2)^3/2. He then differentiated the bottom part and divided by the new power, ultimately multiplying by the differentiation of x^2. This is essentially the same as using the substitution u=x^2+a^2.
  • #1
abdo799
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in this video , the prof had to integrate x/(a^2+x^2)^3/2 , i know we usually do this using substitution , but in the video...he ignored the x and integrate like it was 1/(a^2+x^2)^3/2, how does that work?
 
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  • #2
You will probably need to give more information about what he did.
 
  • #3
he said integration of x/(a^2+x^2)^3/2= x*(-2)/(a^2+x^2)^1/2*2x
i really don't know what he did, he differentiated the bottom part then divided by new power and multiplied by differentiation of x^2
 
  • #5
there was a mistake with the powers in the question and i corrected it
 
  • #6
Look carefully at the integrand x/(a^2+x^2)^(2/3). What is the derivative of (a^2+x^2)? Is it x times some constant perhaps? Can you rewrite the integrand as the product of two expressions, rather than the quotient?

BTW, your video requires a login to view, so we can't see it.
 
  • #7
SteamKing said:
Look carefully at the integrand x/(a^2+x^2)^(2/3). What is the derivative of (a^2+x^2)? Is it x times some constant perhaps? Can you rewrite the integrand as the product of two expressions, rather than the quotient?

BTW, your video requires a login to view, so we can't see it.


the power on the brakets is 3/2
 

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  • #8
abdo799 said:
the power on the brakets is 3/2

The power on the brackets is immaterial. The principle remains.
 
  • #9
SteamKing said:
The power on the brackets is immaterial. The principle remains.

i figured out what he did, he differentiated the bottom part...and that's it
 
  • #10
In other words he did exactly the "[itex]u= x^2+ a^2[/itex]" substitution, just not writing it out explicitly.
 

FAQ: How Can You Integrate x/(a^2+x^2)^(3/2) Without Explicit Substitution?

What is the formula for integration of x/(a^2+x^2)^2/3?

The formula for integration of x/(a^2+x^2)^2/3 is ∫x/(a^2+x^2)^2/3 dx = -1/(3a^2 √(a^2+x^2)) + C, where C is the constant of integration.

What is the significance of the number 'a' in the formula for integration of x/(a^2+x^2)^2/3?

The number 'a' represents the radius of the sphere in which the volume is being calculated. It is the distance from the center of the sphere to the edge of the sphere.

Can the formula for integration of x/(a^2+x^2)^2/3 be used for any value of 'a'?

Yes, the formula can be used for any value of 'a'. However, if 'a' is negative, the result of the integration will also be negative.

Why is the integration of x/(a^2+x^2)^2/3 considered a difficult problem in calculus?

The integration of x/(a^2+x^2)^2/3 is considered difficult because it involves the use of trigonometric substitutions and various integration techniques such as integration by parts or u-substitution. It also requires a deep understanding of the fundamental principles of calculus.

Is there a real-life application of the integration of x/(a^2+x^2)^2/3?

Yes, the integration of x/(a^2+x^2)^2/3 has real-life applications in physics, particularly in calculating the electric field intensity due to a spherical charge distribution. It is also used in calculating the volume of spheres and finding the center of mass of a sphere.

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