# Integration: Fraction With 'x' Variable

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• Jun 13th 2008, 04:11 AM
Simplicity
Integration: Fraction With 'x' Variable
$\displaystyle \int \frac{x}{(x^2-1)^\frac12} \ \mathrm{d}x$

There is something really simple that I am not seeing? (Worried) Help please, Thanks in advance.
• Jun 13th 2008, 04:26 AM
Isomorphism
Quote:

Originally Posted by Air
$\displaystyle \int \frac{x}{(x^2-1)^\frac12} \ \mathrm{d}x$

There is something really simple that I am not seeing? (Worried) Help please, Thanks in advance.

With the substitution $\displaystyle u =x^2 - 1$,

$\displaystyle \int \frac{x}{(x^2-1)^\frac12} \ \mathrm{d}x = \frac12 \int \frac{1}{u^\frac12} \ \mathrm{d}u = \frac12 \int u^{-\frac12} \ \mathrm{d}u$
• Jun 13th 2008, 05:06 AM
colby2152
Quote:

Originally Posted by Air
$\displaystyle \int \frac{x}{(x^2-1)^\frac12} \ \mathrm{d}x$

There is something really simple that I am not seeing? (Worried) Help please, Thanks in advance.

Substitution, let $\displaystyle u = x^2 -1$

Then $\displaystyle du = 2x dx$

Your new integral is $\displaystyle \int 0.5u^{\frac{-1}{2}} du$
• Jun 13th 2008, 06:27 AM
Krizalid
Or make $\displaystyle z^2=x^2-1.$ :(