Partial fractions with fractional powers

[jdstokes]

Homework Statement

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

 

[moo5003]

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

 

[gabbagabbahey]

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

 

[moo5003]

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.)

 

[jdstokes]

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

 

[HallsofIvy]

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

 

[icystrike]

Homework Statement

How does one integrate e.g. \frac{1+x}{(2+x)^{3/2}} 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 \sqrt{x+2} (1+\frac{1}{x+2})

 

[gabbagabbahey]

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

Yup

 

[gabbagabbahey]

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

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.

 

[icystrike]

Yup

hahas. thanks loads for your hint
CHEERS!