Integral of a fraction

**Aperson** · March 8th 2011, 11:27 PM

Hey
So im doing integrals of fractions for my monday testm but theres somehting i dont get about the formula

$integrate(dx/(a^2-u^2)$

I have

However the book says

Why is this?? It took u=(3x+1) and a=sqrt(2) but it inverted them on the logarithm part??

So it ended up using $integrate dx/(u^2-a^2)$ instead..
Thanks for the help ;D

**tonio** · March 9th 2011, 01:59 AM

Originally Posted by Aperson

Hey
So im doing integrals of fractions for my monday testm but theres somehting i dont get about the formula

$integrate(dx/(a^2-u^2)$

I have

However the book says

Why is this?? It took u=(3x+1) and a=sqrt(2) but it inverted them on the logarithm part??

So it ended up using $integrate dx/(u^2-a^2)$ instead..
Thanks for the help ;D

After "I have" and "however the book says" I see nothing, but I guess you meant

$\displaystyle{\int \frac{du}{a^2-u^2}=-\int \frac{du}{u^2-a^2}=-\int\frac{1}{2a}\left(\frac{1}{u-a}-\frac{1}{u+a}\right)\,du=}$

$\displaystyle{-\frac{1}{2a}\ln\left(\frac{u-a}{u+a}\right)+C=\frac{1}{a}\ln\sqrt{\frac{u+a}{u-a}}+C}$

Now, if it really is the integral as you wrote it, then it is even easier:

$\displaystyle{\int\frac{dx}{a^2-u^2}=\frac{x}{a^2-u^2}+C}$

Tonio

**Aperson** · March 9th 2011, 02:53 AM

It was some problem that used those formulas, but those formula equalities you posted where exactly what i was missing lol thanks ;D

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