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

  • Thread starter Thread starter araz1
  • Start date Start date
  • Tags Tags
    Integrate
Click For Summary
SUMMARY

The integral of 3/(x(2 - sqrt(x))) can be solved without using partial fractions by employing the substitution u = sqrt(x). This leads to the transformation of the integral into a more manageable form, specifically 6∫(1/(1 - v^2)) dv, which can be directly integrated. The discussion highlights a method involving a clever substitution to simplify the integral, ultimately resulting in an expression that involves an inverse hyperbolic trigonometric function.

PREREQUISITES
  • Understanding of integral calculus and substitution methods
  • Familiarity with hyperbolic functions and their integrals
  • Knowledge of algebraic manipulation and simplification techniques
  • Experience with integration techniques excluding partial fractions
NEXT STEPS
  • Study the method of integration by substitution in detail
  • Learn about inverse hyperbolic functions and their properties
  • Explore alternative integration techniques beyond partial fractions
  • Practice solving integrals involving square roots and rational functions
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on calculus, as well as educators seeking effective methods for teaching integration techniques without relying on partial fractions.

araz1
Messages
9
Reaction score
0
integral of 3/(x(2-sqrt(x))) using u=sqrt(x)?
But not using partial fractions
 
Physics news on Phys.org
[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
 
Last edited by a moderator:
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?
 

Thread 'How does this derivative work? And why the speed of light remains?'

Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...
View full post »

Similar threads

High School Integration by parts of inverse sine, a solved exercise, some doubts...
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Different Substitution for Solving an Integral
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Problem with calculating projections of curl using rotation of contour
  • · Replies 9 ·
Replies
9
Views
981
Undergrad Substitution in a definite integral
  • · Replies 6 ·
Replies
6
Views
3K
Undergrad How Do You Integrate Expressions Like These?
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad How Do You Integrate 1/√(x^3 + x^2 + x + 1) dx?
  • · Replies 8 ·
Replies
8
Views
2K
MHB Would Partial Fractions Simplify This Integral?
  • · Replies 8 ·
Replies
8
Views
2K
Undergrad Integration of √(1-x^2) with x=sinθ: Check Correctness and Simplification
  • · Replies 12 ·
Replies
12
Views
3K
MHB How Do You Express and Evaluate This Integral Using Partial Fractions?
  • · Replies 6 ·
Replies
6
Views
2K
High School Question about the limits of integration in a change of variables
  • · Replies 2 ·
Replies
2
Views
3K