Integrate 3/(x(2-sqrt(x))) - No Partial Fractions

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Discussion Overview

The discussion centers around the integration of the function 3/(x(2-sqrt(x))) without employing partial fractions. Participants explore various substitution methods and transformations to approach the integral.

Discussion Character

  • Mathematical reasoning, Exploratory

Main Points Raised

  • Some participants propose using the substitution u = sqrt(x) to simplify the integral.
  • One participant transforms the integral into a form involving u and suggests a method involving a trick with variables a and v.
  • Another participant notes that the antiderivative will likely involve an inverse hyperbolic trig function, given the constraints of not using partial fractions.
  • There is a question raised about the rationale behind not using partial fractions in the integration process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the use of partial fractions, with one participant questioning its exclusion while others proceed without it. The discussion remains unresolved regarding the best approach to the integral.

Contextual Notes

Some transformations and substitutions may depend on specific assumptions about the variables involved, and the discussion does not resolve the mathematical steps leading to the proposed antiderivative forms.

araz1
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integral of 3/(x(2-sqrt(x))) using u=sqrt(x)?
But not using partial fractions
 
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[math]\int \dfrac{3}{x(2 - \sqrt{x} )} ~ dx = \int \dfrac{3}{u^2 (2- u)} ~ 2u ~du = \int \dfrac{6}{u (2 - u)} ~du[/math]

Let's pull a trick that occasionally works. I'm going to let 2 - u = a - v and u = a + v. If we add these two equations we get 2 = 2a. Thus a = 1. So 2 - u = 1 - v and u = 1 + v. Putting this into your integral gives
[math]\int \dfrac{6}{u (2 - u)} ~ du = \int \dfrac{6}{(1 + v)(1 - v)} ~ dv = \int \dfrac{6}{1 - v^2} ~ dv[/math]

which you should be able to calculate directly. (Unless you know the trick to integrate this it is going to be difficult.)

-Dan
 
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araz said:
integral of 3/(x(2-sqrt(x))) using u=sqrt(x)?
But not using partial fractions

$u = \sqrt{x} \implies du = \dfrac{dx}{2\sqrt{x}}$

$\displaystyle 3 \int \dfrac{dx}{x(2-\sqrt{x})} = 2 \cdot 3\int \dfrac{dx}{2\sqrt{x}(2\sqrt{x} - x)}$

substitute ...

$\displaystyle 6\int \dfrac{du}{2u - u^2} = -6 \int \dfrac{du}{(u^2 - 2u + 1) - 1} = -6 \int \dfrac{du}{(u-1)^2 - 1}$

since partial fractions is off the table, the antiderivative will involve an inverse hyperbolic trig function
 
topsquark said:
[math]\int \dfrac{3}{x(2 - \sqrt{x} )} ~ dx = \int \dfrac{3}{u^2 (2- u)} ~ 2u ~du = \int \dfrac{6}{u (2 - u)} ~du[/math]

Let's pull a trick that occasionally works. I'm going to let 2 - u = a - v and u = a + v. If we add these two equations we get 2 = 2a. Thus a = 1. So 2 - u = 1 - v and u = 1 + v. Putting this into your integral gives
[math]\int \dfrac{6}{u (2 - u)} ~ du = \int \dfrac{6}{(1 + v)(1 - v)} ~ dv = \int \dfrac{6}{1 - v^2} ~ dv[/math]

which you should be able to calculate directly. (Unless you know the trick to integrate this it is going to be difficult.)

-Dan
Thank you for your time.
 
skeeter said:
$u = \sqrt{x} \implies du = \dfrac{dx}{2\sqrt{x}}$

$\displaystyle 3 \int \dfrac{dx}{x(2-\sqrt{x})} = 2 \cdot 3\int \dfrac{dx}{2\sqrt{x}(2\sqrt{x} - x)}$

substitute ...

$\displaystyle 6\int \dfrac{du}{2u - u^2} = -6 \int \dfrac{du}{(u^2 - 2u + 1) - 1} = -6 \int \dfrac{du}{(u-1)^2 - 1}$

since partial fractions is off the table, the antiderivative will involve an inverse hyperbolic trig function
Thank you for your time.
 
Why wouldn't you use partial fractions though?
 

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