Find Order Notation of x√(1+x^2)

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Teacher has:

x\sqrt{1+x^2}=x+\frac{1}{2}x^3...(= O(x))

in finding order notation of expression.

How is L.H.S. equal to R.H.S.?
 
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The first equation is an expansion of the function x\sqrt{1 + x^2} in a Taylor series at zero, using the binomial series.

The second equation is a statement that the series expansion is bounded by a constant multiple of x. This is true only near zero, not near infinity (near infinity, \sqrt{1 + x^2} looks like |x|, so x\sqrt{1 + x^2} grows quadratically). Near zero, the written-out version of the big-O notation is: there exist constants C > 0 and \delta > 0 so that, whenever |x| < \delta, |x\sqrt{1 + x^2}| < C|x|. This statement is abbreviated x\sqrt{1 + x^2} = O(x) \textrm{ as } x \to 0.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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