# Integrating Factor

• April 18th 2009, 01:13 PM
Macleef
Integrating Factor
I don't get the step in the red box.

But.. here's what I have:
$
I = e^{ \int \frac{x}{x^2 + 5} \, dx}$

$I = e^{\int x \frac{1}{x^2 + 5} \, dx}$

$I = e^{\frac{1}{2}x^2 + ln(x^2 + 5)}$

$I = e^{\frac{1}{2}x^2}e^{ln(x^2 + 5)}$

$I = e^{x}(x^2 + 5)$
• April 18th 2009, 01:16 PM
Moo
Hi,

and what you did is wrong o.O $\int fg \neq \int f \int g$
• April 18th 2009, 01:22 PM
Macleef
Quote:

Originally Posted by Moo
Hi,

and what you did is wrong o.O $\int fg \neq \int f \int g$

I did that but I still don't know what to do...

$u = x^2 + 5$

$du = 2x + 5 \, dx$

$I = e^{\int (x)(\frac{1}{u}(\frac{1}{2x + 5}) \, du}$

How do I cancel the "x"s if there's an addition at the denominator ??
• April 18th 2009, 01:27 PM
Moo
Quote:

Originally Posted by Macleef
I did that but I still don't know what to do...

$u = x^2 + 5$

$du = 2x + 5 \, dx$

$I = e^{\int (x)(\frac{1}{u}(\frac{1}{2x + 5}) \, du}$

How do I cancel the "x"s if there's an addition at the denominator ??

What is the derivative of x^2+5 with respect to x ??
What is the derivative of a constant ?

du is not (2x+5) dx.

I'm sorry, I'll let you search for your mistake, I won't tell you the solution (Evilgrin)
• April 18th 2009, 01:38 PM
Macleef
wow... I've been doing so much differentiation that I almost forgot all basic rules of derivatives. Thank you!
• April 18th 2009, 01:40 PM
Moo
Quote:

Originally Posted by Macleef
wow... I've been doing so much differentiation that I almost forgot all basic rules of derivatives. Thank you!

That's less scary ! (Giggle)