Thread: Calc 1 integral

1. Calc 1 integral

I am having trouble figuring out how to solve integrals that are rational, and also when square roots are involved. Please bear with me, I'm in an online class and I've discovered that I'm not very good at teaching myself

Evaluate the integral:
(s(s+1)^2 / sqrt(s))ds

what is the rule of thumb in approaching a situation like this? I've been trying to think backwards to see what would have to exist to end up with this as a derivative, but every time the product rule leaves me with a lot more than what is here.

help? thanks!

2. Exapnd it out and integrate term by term. Then, it's rather easy.

$\frac{s(s+1)^{2}}{\sqrt{s}}=s^{\frac{5}{2}}+2s^{\f rac{3}{2}}+s^{\frac{1}{2}}$

3. Okay, that makes sense. That way all of the terms in the numerator have an "s" and then the denominator can go away.

But would the same idea apply in this example? Or is there another trick that can be used?

evaluate the integral:

3y / (2y^2 + 5)^1/2

I've tried to expand the denominator, but I don't see how I can get a "y" in all terms.

What is the process here? I would love to know how to make it as easy as the last one!

4. Originally Posted by littlejodo
Okay, that makes sense. That way all of the terms in the numerator have an "s" and then the denominator can go away.

But would the same idea apply in this example? Or is there another trick that can be used?

evaluate the integral:

3y / (2y^2 + 5)^1/2

I've tried to expand the denominator, but I don't see how I can get a "y" in all terms.

What is the process here? I would love to know how to make it as easy as the last one!
This one you canot expand but notice that if you let $m=2y^2+5$

Then this integral transforms into $\frac{3}{4}\int\frac{dm}{\sqrt{m}}$

5. thanks! I don't know if I will ever be able to see those connections on my own. I think my brains just isn't used to it