# Partial fractions with fractional powers

1. Nov 26, 2008

### jdstokes

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

2. Nov 26, 2008

### moo5003

This looks interesting, I have never done it but my guess would be to make a substitution.

3. Nov 26, 2008

### gabbagabbahey

Hint: $x+1=(x+2)-1$

4. Nov 26, 2008

### moo5003

Nice catch gabba, cant believe I didnt see that ^_^. Though, does that satisfy it being a partial fraction after you simplify/reduce? (Looks up the def.)

5. Nov 26, 2008

### jdstokes

Yes, nice one. Is splitting the numerator in this way a special case of partial fractions? Not that it really matters....

6. Nov 27, 2008

### HallsofIvy

Staff Emeritus
No. That is NOT using "partial fractions". This can be integrated but not by partial fractions.

7. Nov 27, 2008

### icystrike

is the ans
2 $$\sqrt{x+2}$$ (1+$$\frac{1}{x+2}$$)

Last edited: Nov 27, 2008
8. Nov 27, 2008

### gabbagabbahey

Yup

9. Nov 27, 2008

### gabbagabbahey

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

Either way, it sure seems like "partial fraction decomposition" to me.

10. Nov 27, 2008