Basic question about an Integration example

**Lancet** · March 4th 2011, 03:47 PM

I understand the basics of integration, but I'm extremely rusty, and in the process of refreshing myself, I hit an example I don't understand.

This is the problem:

$ \int \frac{2x}{1+x^2}dx $

So, my understanding of the correct way to solve this is this:

$ 2\int \frac{x}{1+x^2}dx $

$ u = 1+x^2 $

$ du = 2xdx $

$ =\int \frac{1}{u}du $

$ =ln|u|+c $

$ =ln|1+x^2|+c $

Here's what I don't understand - what happened to $du$ ? $2xdx$ ?

To me, it just dissappeared.

**e^(i*pi)** · March 4th 2011, 03:59 PM

$2x\, dx$ was substituted for the equivalent expression of $du$ which is why you used your sub.

It's the same principle of $x+y = 2 \implies x+3x=2$ given that $y=3x$

**Lancet** · March 4th 2011, 04:14 PM

That's what I would have thought, but wouldn't that have left us with this?

$ ln|u|(2x) $

**e^(i*pi)** · March 4th 2011, 04:44 PM

No, the 2x is part of the expression which is being subbed so it goes too

Rearrange to $dx = \dfrac{du}{2x}$ and sub in if it helps

**Prove It** · March 4th 2011, 09:57 PM

Rewrite the integral as

$\displaystyle \int{\left(\frac{1}{1 + x^2}\right)\,2x\,dx}$ .

If you let $\displaystyle u = 1 + x^2$ then $\displaystyle \frac{du}{dx} = 2x$ and the integral becomes

$\displaystyle \int{\frac{1}{u}\,\frac{du}{dx}\,dx}$

$\displaystyle = \int{\frac{1}{u}\,du}$ . (This is an application of the Chain Rule in reverse).

**TheCoffeeMachine** · March 4th 2011, 11:09 PM

Also if you know that $\int\frac{f'(x)}{f(x)}\;{dx} = \ln{f(x)}+k$ , then
you should be able to write down the answer with no work.
(Even if you don't, it's a useful form to learn and remember).

**Prove It** · March 4th 2011, 11:24 PM

Originally Posted by TheCoffeeMachine

Also if you know that $\int\frac{f'(x)}{f(x)}\;{dx} = \ln{f(x)}+k$ , then
you should be able to write down the answer with no work.
(Even if you don't, it's a useful form to learn and remember).

Actually $\displaystyle \int{\frac{f'(x)}{f(x)}\,dx} = \ln{|f(x)|} + k$ ... And this is a standard result of substitution

**TheCoffeeMachine** · March 4th 2011, 11:49 PM

Originally Posted by Prove It

Actually $\displaystyle \int{\frac{f'(x)}{f(x)}\,dx} = \ln{|f(x)|} + k$ ... And this is a standard result of substitution

I'm in the habit of leaving out the signs for some reason. :)

**tom@ballooncalculus** · March 5th 2011, 01:03 AM

Just in case a picture helps to follow your way back through the chain rule...

... where (key in spoiler) ...

Spoiler:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

**Lancet** · March 5th 2011, 06:29 AM

Originally Posted by Prove It

Rewrite the integral as

$\displaystyle \int{\left(\frac{1}{1 + x^2}\right)\,2x\,dx}$ .

If you let $\displaystyle u = 1 + x^2$ then $\displaystyle \frac{du}{dx} = 2x$ and the integral becomes

$\displaystyle \int{\frac{1}{u}\,\frac{du}{dx}\,dx}$

$\displaystyle = \int{\frac{1}{u}\,du}$ . (This is an application of the Chain Rule in reverse).

That helped!. I think I'm starting to grasp it now. Part of what I didn't understand is that if $\frac{du}{dx}$ is coming from differentiating $u$ , where did the original $2x$ go. But I think I understand that that was being swapped for the $2x$ due to the equality, rather than being inserted.

It also explains the second part of what I didn't understand (but didn't post) which is why, if the numerator was $4x$ did the answer equal this one, but with a $2$ as a multiplier in front of everything.

I was using Wolfram Alpha to show me the steps, but the problem was that it wasn't showing me enough of the intermediate steps for me to fully grasp it.

Thank you! And thanks to everyone else who tried to knock this into my brain as well.

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