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

• MHB
• araz1
In summary, the conversation discusses how to integrate 3/(x(2-sqrt(x))) using the substitution u=sqrt(x) and avoiding the use of partial fractions. The expert suggests using a trick of letting 2-u = a-v and u = a+v to simplify the integral, but ultimately the antiderivative will involve an inverse hyperbolic trig function.
araz1
integral of 3/(x(2-sqrt(x))) using u=sqrt(x)?
But not using partial fractions

$$\displaystyle \int \dfrac{3}{x(2 - \sqrt{x} )} ~ dx = \int \dfrac{3}{u^2 (2- u)} ~ 2u ~du = \int \dfrac{6}{u (2 - u)} ~du$$

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

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:
$$\displaystyle \int \dfrac{3}{x(2 - \sqrt{x} )} ~ dx = \int \dfrac{3}{u^2 (2- u)} ~ 2u ~du = \int \dfrac{6}{u (2 - u)} ~du$$

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

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

-Dan

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

Why wouldn't you use partial fractions though?

## What is the general process for integrating 3/(x(2-sqrt(x))) without using partial fractions?

The general process for integrating this function is to first rewrite it as a sum of simpler fractions using the decomposition method. Then, use the substitution method to change the variable in the integral to make it easier to solve. Finally, integrate each term separately and combine the results to get the final solution.

## What is the decomposition method and how does it help in integrating this function?

The decomposition method involves breaking down a complex fraction into simpler fractions. In the case of 3/(x(2-sqrt(x))), we can rewrite it as 3/(x(2-sqrt(x))) = A/x + B/(2-sqrt(x)), where A and B are constants. This helps in integrating the function because it allows us to split the integral into smaller, more manageable parts.

## Why do we need to use substitution when integrating this function?

Substitution allows us to change the variable in the integral to make it easier to solve. In the case of 3/(x(2-sqrt(x))), we can use the substitution u = 2-sqrt(x) to change the variable from x to u. This simplifies the integral and makes it easier to integrate each term separately.

## What are the steps for using the substitution method in this integral?

The steps for using the substitution method in this integral are as follows: 1) Identify a suitable substitution, 2) Rewrite the integral in terms of the new variable, 3) Calculate the derivative of the new variable, 4) Substitute the new variable and its derivative into the integral, 5) Integrate each term separately, and 6) Substitute the original variable back into the solution.

## Is there a simpler way to solve this integral without using partial fractions?

No, using partial fractions is the most efficient and straightforward way to solve this integral. Other methods may involve more complex algebraic manipulations and may not always work for all integrals. The decomposition and substitution method used in this approach is the most efficient and reliable way to solve this integral.

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