Question

I'm having trouble understanding an intermediate step in a solution of an indefinite integral, can somebody help me? (Integral below in a minute).

Answer 1

\[\int\limits_{?}^{?} \frac{3 }{ x + 2\sqrt{x} }\]

Answer 2

First, the constant of 3 is factored outside the integral, and then after that, by u-substitution, they get this:

Answer 3

\[6 \int\limits_{}^{} \frac{ u }{ u ^{2}+u }du\]

Answer 4

What I don't understand is why the constant (3) doubled, and how, when using u-substitution where u is (sqrt)x and du is 1/2(sqrt)x du, why the numerator of the substituted integral is u.

Answer 5

Let me clarify my thought process by showing what I would do, and then maybe somebody can point out where I'm wrong, give me a moment:

Answer 6

\[\int\limits_{}^{}\frac{ 3 }{ x+2\sqrt{x} } = 3\int\limits_{}^{}\frac{ 1 }{ x+2\sqrt{x} }\]
\[u = \sqrt{x}, du = \frac{ 1 }{ 2\sqrt{x} }dx\]
\[3\int\limits_{}^{}\frac{ 1 }{ u ^{2}+2u }du\]

I would have ended up like that. I don't know what's wrong with my substitution. I don't understand how they're getting a u in their numerator, or why the constant factored out changes from three to six.

Answer 7

To me it seems that you have a recipe to 'simplify' the integral before trying to solve it. Forget about the constant 3 (or 6) for a moment and focus on the replacement of x. As there is a root(x) in your formula, you would like to find an easy replacement which removes this root(x), which looks too scary to work with.

The best choice is then: replace x by u^2, because root(x) will then be replaced by root(u^2) which happens to be just 'u'.

In your original problem, there also needs to be a dx at the end of your integral. Since you have replaced x by u^2, you also need to replace dx by d(u^2), which is 2*du. This is where the 2 is introduced on the top part of your integral. Together with the 3, which was already there, you end up with 6 on the topside. This '6' can be brought before the integral sign as you already captured in your posting. Hope this is clear now ?

Answer 8

Yep, in your approach you didn't include the dx in the original problem.

Answer 9

I do it and get everything except u^2 +u. I get the 6 and the u on top, but I get u^2 + 2u on bottom, lol. I have an extra 2 from what he said the answer was.

Answer 10

Sorry, I forget the dx's. They're obviously important but I have a habit of it seeming like a pedantic thing tacked on the bottom. One second, I'm having a little trouble understanding your explanation, NL, but thanks, and give me a minute.

Answer 11

I think NL pointed it out (and not sure if you do it), but it may help to solve for dx when doing u-sub. I always take an extra moment to solve for it and sub it back into the problem and then simplify.

Answer 12

Yea, what you typed was an error. Where you put u^2 + u in the denominator, it's supposed to be u^2 + 2u.

Answer 13

Oh, got it! I got the answer, but I came to it in a slightly different way that still works repeatedly.