<s>[INST] Can i factor out an $x^2$ while completing the square? Original equation $$\int\frac{x}{\sqrt{56+10x^2-x^4}}$$ 
This equation must be integrated by completing the square. My question is, when completing the square am i able to factor out $x^2$ like this $$-x^4+10x^2+56$$ $$-x^2(x^2-10)+56$$ would that be fine? and if is so how would I continue on about this with the $-x^2$ on the outside?
 [/INST] $$\int\frac{x}{\sqrt{56+10x^2-x^4}}dx$$
Using the substitution $x^2 = t$, $\,\,dt = 2xdx$
$$\frac{1}{2}\int\frac{dt}{\sqrt{-(t^2 -10t - 56)}}$$
$$\frac{1}{2}\int\frac{dt}{\sqrt{81-(t-5)^2}}$$ 
Substitute $s = t-5$, $\,\,ds = dt$
$$\frac{1}{2}\int\frac{ds}{\sqrt{81-s^2}}$$
$$\frac{1}{2}\int\frac{ds}{9\sqrt{1-\frac{s^2}{81}}}$$
Substitute $p = \frac{s}{9}$, $\,\,dp = \frac{ds}{9}$
$$=\frac{1}{2}\int\frac{dp}{\sqrt{1-p^2}}$$
$$=\frac{1}{2}\sin^{-1}(p)$$
$$=\frac{1}{2}\sin^{-1}\Big(\frac{s}{9}\Big)$$
$$=\frac{1}{2}\sin^{-1}\Big(\frac{t-5}{9}\Big)$$
$$=\frac{1}{2}\sin^{-1}\Big(\frac{1}{9}(x^2-5)\Big)$$
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