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

[araz1]

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

 

[topsquark]

[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

 

[skeeter]

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

 

[araz1]

[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.

 

[araz1]

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

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

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.

 

[Prove It]

Why wouldn't you use partial fractions though?