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

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Homework Statement



My solution has two terms divided by a which is in error. I am a volunteer math coach to some junior college students and can't find my error in the problem. Its been a while since I earned my masters in electrical engineering.

The issue is the presencce of the variable "a" in the denominator of two of the terms in the answer. I can't imagine why the "a"s are not in the correct answer.

Homework Equations


The problem requires solution by trigonometric substitution. All other equations are listed in my attempt at the problem.

The Attempt at a Solution


My work is all on the solution attempt.
 

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Seems like an ##x=atan \theta## or ##x=asec \theta ## would work.
 
WWGD said:
Seems like an ##x=atan \theta## or ##x=asec \theta ## would work.

The solution I provided attempts to use x = a tan theta but the solution is incorrect
 
Ray Beaver said:
The solution I provided attempts to use x = a tan theta but the solution is incorrect

Is the solution an indefinite (general) or definite integral (numerical value)?

EDIT: Nevermind, I did not read carefully and did not see the 3/2 exponent. Please ignore and let me
rethink. It is a variant of the one with power 1/2, but needs some changes.
 
Last edited:
The solution is listed at the bottom of the attempt I posted and is indefinite.
 
The solution at the bottom of the solution page is missing a minus sign in front of the first term, my mistake in listing the correct answer. My answer has that term correct, but is in error in the argument of the natural log.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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