# Integral from apostol

1. Sep 23, 2015

### Ted13

1. The problem statement, all variables and given/known data
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∫xn-1⋅(ax+b)½)

2. Relevant equations

3. The attempt at a solution
Setting u=xn,dv=(ax+b)1/2dx
du=n⋅xn-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

2. Sep 23, 2015

### Staff: Mentor

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.

3. Sep 23, 2015

### Ted13

Thank's Setting u=(ax+b) du=a⋅dx then the integral becomes ∫u1/2du=1/a⋅u3/2⋅2/3 but that's not enough for proving the whole thing

4. Sep 23, 2015

### Staff: Mentor

∫xn⋅(ax+b)½=2/a(2n+3)(xn⋅(ax+b)3/2-nb∫xn-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 $\frac{2}{a(2n + 3)}$?

5. Sep 23, 2015

### Ted13

It is 2 over a(2n+3)

6. Sep 23, 2015

### Staff: Mentor

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.

7. Sep 23, 2015

### Ted13

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

8. Sep 23, 2015

### Staff: Mentor

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...

9. Sep 24, 2015

### Ted13

Any ideas???

10. Sep 24, 2015

### Staff: Mentor

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.