Partial fractions with fractional powers

In summary, the conversation discusses how to integrate a specific function using partial fractions. One person is unsure about the technique, while another suggests making a substitution and using partial fraction decomposition. The final answer is confirmed to be using partial fractions.
  • #1
jdstokes
523
1

Homework Statement



How does one integrate e.g. [itex]\frac{1+x}{(2+x)^{3/2}}[/itex] by partial fractions?

The Attempt at a Solution



I have no idea about this. I've never seen this technique applied with fractional powers before.
 
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  • #2
This looks interesting, I have never done it but my guess would be to make a substitution.
 
  • #3
Hint: [itex]x+1=(x+2)-1[/itex]:wink:
 
  • #4
Nice catch gabba, can't believe I didnt see that ^_^. Though, does that satisfy it being a partial fraction after you simplify/reduce? (Looks up the def.)
 
  • #5
Yes, nice one. Is splitting the numerator in this way a special case of partial fractions? Not that it really matters...
 
  • #6
No. That is NOT using "partial fractions". This can be integrated but not by partial fractions.
 
  • #7
jdstokes said:

Homework Statement



How does one integrate e.g. [itex]\frac{1+x}{(2+x)^{3/2}}[/itex] by partial fractions?

The Attempt at a Solution



I have no idea about this. I've never seen this technique applied with fractional powers before.

is the ans
2 [tex]\sqrt{x+2}[/tex] (1+[tex]\frac{1}{x+2}[/tex])
 
Last edited:
  • #8
icystrike said:
is the ans
2 [tex]\sqrt{x+2}[/tex] (1+[tex]\frac{1}{x+2}[/tex])

Yup :smile:
 
  • #9
HallsofIvy said:
No. That is NOT using "partial fractions". This can be integrated but not by partial fractions.

Sure it is: if you let [itex]u=(2+x)^{1/2}[/itex] then both the numerator and denominator are polynomials in powers of [itex]u[/itex]. You can the decompose it into partial fractions by letting [tex]\frac{1+x}{(2+x)^{3/2}}=\frac{u^2-1}{u^3}=A+\frac{B}{u}+\frac{C}{u^2}+\frac{D}{u^3}[/tex] and determining A,B,C and D...or you can simply recognize that the fraction decomposes into [tex]\frac{1}{(2+x)^{1/2}}-\frac{1}{(2+x)^{3/2}}[/tex]

Either way, it sure seems like "partial fraction decomposition" to me.
 
  • #10
gabbagabbahey said:
Yup :smile:

hahas. thanks loads for your hint :smile:
CHEERS!
 

1. What are partial fractions with fractional powers?

Partial fractions with fractional powers are expressions that involve fractions with fractional exponents. They are typically used in integration problems to simplify the integration process.

2. How do I solve partial fractions with fractional powers?

To solve partial fractions with fractional powers, you must first decompose the fraction into simpler fractions using the method of partial fractions. Once the fraction is decomposed, you can integrate each individual fraction separately.

3. What is the method of partial fractions?

The method of partial fractions is a technique used to break down a complex fraction into simpler fractions. It involves finding the partial fraction decomposition, which is a representation of the original fraction as a sum of simpler fractions.

4. When are partial fractions with fractional powers used?

Partial fractions with fractional powers are commonly used in integration problems involving rational functions. They allow for easier integration and can provide more accurate solutions.

5. Are there any special rules for solving partial fractions with fractional powers?

Yes, there are some rules to follow when solving partial fractions with fractional powers. These include making sure the degree of the numerator is less than the degree of the denominator, and handling repeated and complex roots appropriately.

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