Part of an Indefinite Integral Problem

**!!!** · March 16th 2008, 08:53 PM

How do I get from the left side of this equation to the right?

**roy_zhang** · March 16th 2008, 09:01 PM

It looks like just applying the quotient rule.

**!!!** · March 16th 2008, 09:10 PM

Sorry, I should have added more detail to my first post. I know that I should apply the quotient rule, but my answer keeps differing from the correct one above. Can somebody please show the the work so I know what I'm doing wrong?

**roy_zhang** · March 16th 2008, 09:29 PM

Ok, let's see.

$\frac{d}{dx}[\frac{\sqrt{x^2+a^2}}{x}]=\frac{(x)\frac{d}{dx}[\sqrt{x^2+a^2}]-\sqrt{x^2+a^2}(\frac{d}{dx}x)}{x^2}$
$=\frac{(x)[\frac{1}{2}(x^2+a^2)^{(-\frac{1}{2})}(2x)]-\sqrt{x^2+a^2}\cdot 1}{x^2}=\frac{x(\frac{x}{\sqrt{x^2+a^2}})-\sqrt{x^2+a^2}\cdot 1}{x^2}$

Then multiply the fraction with the coefficient $-\frac{1}{a^2}$ , you will have the desired result.

Roy

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