# Basic question about an Integration example

Printable View

• March 4th 2011, 03:47 PM
Lancet
Basic question about an Integration example
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.
• March 4th 2011, 03:59 PM
e^(i*pi)
$2x\, dx$ 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 $y=3x$
• March 4th 2011, 04:14 PM
Lancet
That's what I would have thought, but wouldn't that have left us with this?

$
ln|u|(2x)
$
• March 4th 2011, 04:44 PM
e^(i*pi)
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
• March 4th 2011, 09:57 PM
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).
• March 4th 2011, 11:09 PM
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).
• March 4th 2011, 11:24 PM
Prove It
Quote:

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 :)
• March 4th 2011, 11:49 PM
TheCoffeeMachine
Quote:

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. :)
• March 5th 2011, 01:03 AM
tom@ballooncalculus
Just in case a picture helps to follow your way back through the chain rule...

http://www.ballooncalculus.org/draw/...in/fifteen.png

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

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule). So they replace dx and du.

The general drift is...

http://www.ballooncalculus.org/asy/maps/intChain.png

... but you can also follow it through downwards, to check that your integral differentiates to give the integrand.

http://www.ballooncalculus.org/asy/maps/diffChain.png

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• March 5th 2011, 06:29 AM
Lancet
Quote:

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. :)