How Do You Integrate 1/(x^(1/2) - x^(1/3))?

In summary, the student attempted to solve a homework equation involving u = x1/6 by replacing the integral's variable with u = x1/6. However, the results were not as expected and the student needed help from the tutor to finish.
  • #1
Footballer010
14
0

Homework Statement


[tex]\int\frac{dx}{x^{1/2}-x^{1/3}}[/tex]


Homework Equations



None

The Attempt at a Solution



What I did was replace [tex]\int\frac{dx}{x^{1/2}-x^{1/3}}[/tex] with [tex]\int\frac{dx}{u^{3}-u^{2}}[/tex] if u=x[tex]^{1/6}[/tex]. Just to simplify things. And I think that was pointless. Help me out?
 
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  • #2


I would suggest that you continue with u = x1/6. What was the last thing you did?
 
  • #3


I used integration by parts and got [tex]\frac{1}{u^{2}(u-1)}[/tex]=[tex]\frac{-1}{u}[/tex]-[tex]\frac{1}{u^{2}}[/tex]+[tex]\frac{1}{u-1}[/tex]. Is this right?
 
  • #4


Well, first you need to convert the dx to an expression involving du. To do so, start by taking the derivative of [itex]u = x^{1/6}[/itex].
 
  • #5


So du=[tex]\frac{x^{-5/6}}{6}[/tex]. Now I'm lost again.
 
  • #6


Since you don't have x-5/6 (ignoring the constant 1/6) in the integrand, use u = x1/6 to substitute into du = x-5/6/6 dx (don't forget the dx!) so that you only have the variable u in the integrand.
 
  • #7


u=x[tex]^{1/6}[/tex] so u[tex]^{6}[/tex]=x

du=((u[tex]^{6}[/tex])[tex]^{-5/6}[/tex])/6 dx

So, du=(u[tex]^{-5}[/tex])/6 dx

Now what?
 
  • #8


You want to get dx by itself, then substitute what it's equal to into your integral. After that you shouldn't have any trouble with it.
 
  • #9


Hmm...I entered the integral into my calc and its not even close to what I got. This makes no sense!
 
  • #10


Did you use Bohrok's suggestion? Solve for dx in terms of u and du and substitute it into the integral, so that the integral is entirely in terms of u and du. In other words, what is [itex]f(u)[/itex] in this expression?
[tex]\int\frac{\mathrm{d}x}{x^{1/2}-x^{1/3}} = \int f(u) \mathrm{d}u[/tex]
 
  • #11


I just got it...

6[tex]\int[/tex][tex]u^{5}/(u^{2}(u-1)[/tex])du.

Then I simplified and used long division to get a long complicated answer, but its right. Thank you both!
 
  • #12


You did good on the start.
[tex]
\int\frac{dx}{u^{3}-u^{2}}
[/tex] where [tex]u=x^{1/6}[/tex]

The next step is:

[tex]\int\frac{dx}{u^2(u-1)}[/tex]

What would you do next?

Partial fractions. Then substitution, and finally substitution for x=u^6
 

FAQ: How Do You Integrate 1/(x^(1/2) - x^(1/3))?

What is the definition of "Integral of a root polynomial"?

The integral of a root polynomial is a mathematical expression that represents the area under the curve of a polynomial function with a root or roots. It is the reverse process of differentiation and is used to determine the original function when its derivative is known.

How is the integral of a root polynomial calculated?

The integral of a root polynomial can be calculated using various methods, such as the substitution method, the integration by parts method, or the partial fraction method. The specific method used depends on the complexity of the polynomial function and the availability of known integration techniques.

What is the significance of the integral of a root polynomial?

The integral of a root polynomial has various applications in mathematics, physics, and engineering. It is used to calculate the area under a curve, which is essential in determining the total distance traveled, the total displacement, or the total work done by a variable force. It is also used to solve differential equations, which are used to model various real-life phenomena.

How do the roots of a polynomial affect its integral?

The roots of a polynomial function have a significant impact on its integral. If the polynomial has a root at a particular point, the integral will have a discontinuity at that point. If the polynomial has multiple roots, the integral will have multiple discontinuities. Additionally, the location and multiplicity of the roots can affect the convergence or divergence of the integral.

Can the integral of a root polynomial be negative?

Yes, the integral of a root polynomial can be negative. This occurs when the function has a negative value over a certain interval, resulting in a negative area under the curve. The negative value of the integral is an indication that the function is decreasing over that interval. However, the definite integral of a root polynomial over its entire domain will always be positive or zero.

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