Proving ∫xn⋅(ax+b)½ using Integration by Parts | Integral from Apostol

[Ted13]

Homework Statement

Hi guys,can anyone help me prove the following using integration by parts
∫xⁿ⋅(ax+b)^½=2/a(2n+3)(xⁿ⋅(ax+b)^3/2-nb∫x^n-1⋅(ax+b)^½)

Homework Equations

The Attempt at a Solution

Setting u=xⁿ,dv=(ax+b)^1/2dx
du=n⋅x^n-1 and v = 2/3⋅(ax+b)^3/2 i can never get b in the equation and the term 2n+3 in the denominator

 

[Mark44]

Homework Statement

Hi guys,can anyone help me prove the following using integration by parts
∫xⁿ⋅(ax+b)^½=2/a(2n+3)(xⁿ⋅(ax+b)^3/2-nb∫x^n-1⋅(ax+b)^½)

Homework Equations

The Attempt at a Solution

Setting u=xⁿ,dv=(ax+b)^1/2dx
du=n⋅x^n-1 and v = 2/3⋅(ax+b)^3/2 i can never get b in the equation and the term 2n+3 in the denominator

Your work in finding v is incorrect. When you calculate $\int dv = \int (ax + b)^{1/2} dx$, there's a simple substitution that you need to use.

 

[Ted13]

Thank's Setting u=(ax+b) du=a⋅dx then the integral becomes ∫u^1/2du=1/a⋅u^3/2⋅2/3 but that's not enough for proving the whole thing

 

[Mark44]

Please clarify what this means:
∫xⁿ⋅(ax+b)^½=2/a(2n+3)(xⁿ⋅(ax+b)^3/2-nb∫x^n-1⋅(ax+b)^½)
The first part just to the right of = is ambiguous -- this part: 2/a(2n+3).

Is this $\frac 2 a (2n + 3)$ or is it $\frac{2}{a(2n + 3)}$?

 

[Ted13]

It is 2 over a(2n+3)

 

[Mark44]

It is 2 over a(2n+3)

Then you should write it as 2/(a(2n + 3)). As you wrote it, most people would interpret it as $\frac 2 a (2n + 3)$.

I'll take another look at your integral and see if I can come up with something. BTW, you are omitting the dx differentials in all of your integrals. That's not so bad in this problem, but leaving them off can come back to bite you when you are doing trig substitutions.

 

[Ted13]

I am sorry I'll fix that and thank's for the help again

 

[Mark44]

I've filled up three pages of paper, both sides, but haven't gotten anywhere. I think that the strategy is to do integration by parts twice, but I'm not sure. I need to do some other stuff today, but I'll take another look later on.

Maybe someone else will have an idea...

 

[Ted13]

Any ideas?

 

[mfb]

Setting u=xⁿ,dv=(ax+b)^1/2dx
du=n⋅x^n-1 and v = 2/3⋅(ax+b)^3/2 i can never get b in the equation and the term 2n+3 in the denominator

That is a good approach.

After partial integration, write $v=\frac{2}{3}(ax+b) (ax+b)^{1/2}$ and split the integrand in two summands. You'll get one integral that is proportional to the one on the left side (=the original problem), and one integral that looks like the one you need for your solution. Simplify the whole equation and you are done.