$\displaystyle \int \frac{x}{(x^2-1)^\frac12} \ \mathrm{d}x$ There is something really simple that I am not seeing? Help please, Thanks in advance.
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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? 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 $
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? Help please, Thanks in advance. Substitution, let $\displaystyle u = x^2 -1$ Then $\displaystyle du = 2x dx$ Your new integral is $\displaystyle \int 0.5u^{\frac{-1}{2}} du$
Or make $\displaystyle z^2=x^2-1.$
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