Integrating Difficult Equations: U Substitution

In summary, the conversation discusses the integral of a function involving a radical with a fraction inside, specifically \int \sqrt{\frac{9}{4}x+1}. The question is whether to use u substitution or not, and the correct approach is to use u=\frac{9}{4x}+1. There is some confusion in the conversation about the expression in the radical, but eventually the correct substitution is made and the integral can be easily solved.
  • #1
Stratosphere
373
0

Homework Statement


As I was reviewing some of my previuosly learned calculus I came across somthing that I had either forgoten how to do or was never taught. How do you take the integral of somthing like this

[tex]\int \sqrt{\frac{9}{4}x+1}[/tex]

I don't know how to start with this one. Do I use U substitution?
 
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  • #2
First rewrite the integral as

dy=[tex]\int(\frac{9}{4x}+1)^{1/2}[/tex] dx

Now use the u substitution.

Thanks
Matt
 
  • #3
Using U substitution I got ((9/4x+1)^1/2)/(27/8) I don't think I did this right.
 
  • #4
I don't believe that is correct.

Try this.

Let u = [tex]\frac{9}{4x}[/tex]+1

Now find du/dx.
 
  • #5
Du/dx is =36/16x[tex]^{2}[/tex]
 
  • #6
Can you show your steps as to how you got that result please?
 
  • #7
u=[tex]\frac{9}{4x}+1 [/tex]

[tex]\frac{0*4x-9*4}{(4x)^{2}}[/tex] So I made a mistake with the negative? The answer should be -36/16x^2.
 
  • #8
Yes, you got it. Now perform the rest of the u substitution procedure and your done.
 
  • #9
Stratosphere said:

Homework Statement


As I was reviewing some of my previuosly learned calculus I came across somthing that I had either forgoten how to do or was never taught. How do you take the integral of somthing like this

[tex]\int \sqrt{\frac{9}{4}x+1}[/tex]

I don't know how to start with this one. Do I use U substitution?

I think the discussion got off track somewhere. The radicand above is [itex] \frac{9}{4}x+1[/itex] not [itex]\frac{9}{4x}+1[/itex]. Or was the original statement an error?

If the original is correct, then it can be easilly integrated by substitution - try [itex]u=\frac{9}{4}x+1[/itex].

(This looks like a Stewart's Calculus exercise.)

--Elucidus
 

What is U substitution and why is it useful in integrating difficult equations?

U substitution is a technique used in calculus to simplify the integration of complex functions. It involves replacing a portion of the equation with a new variable, usually denoted as "u", in order to transform the equation into a simpler form that can be integrated more easily. U substitution is useful because it can make the integration process more manageable and often leads to a more elegant solution.

How do I know when to use U substitution?

U substitution is most commonly used when integrating functions with nested functions, such as exponential or trigonometric functions. Another indicator that U substitution may be useful is when the derivative of a function within the equation is also present. It is important to also consider other integration techniques, such as integration by parts, before deciding to use U substitution.

What are the steps for using U substitution?

The steps for using U substitution are as follows:1. Identify the portion of the equation that can be replaced with a new variable, u.2. Substitute the new variable into the equation.3. Calculate the derivative of the new variable, du.4. Rewrite the equation in terms of u and du.5. Integrate the new equation with respect to u.6. Substitute u back into the final solution.

Can U substitution be used for all types of equations?

No, U substitution is not always the most efficient or effective method for integrating all types of equations. It is best suited for functions with nested functions or when the derivative of a function is also present. Other integration techniques, such as integration by parts or trigonometric substitution, may be more appropriate for other types of equations.

Are there any common mistakes to avoid when using U substitution?

One common mistake to avoid is forgetting to include the derivative, du, when rewriting the equation in terms of u. This can lead to incorrect solutions. It is also important to be careful with the limits of integration when substituting u back into the final solution. Another mistake to avoid is using U substitution unnecessarily when a simpler integration technique could be used instead.

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