Can i factor out an $x^2$ while completing the square?

Question

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?

Answer 1

$$\int\frac{x}{\sqrt{56+10x^2-x^4}}dx$$ 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 $p = \frac{s}{9}$, $\,\,dp = \frac{ds}{9}$ $$=\frac{1}{2}\int\frac{dp}{\sqrt{1-p^2}}$$ $$=\frac{1}{2}\sin^{-1}(p)$$ $$=\frac{1}{2}\sin^{-1}\Big(\frac{s}{9}\Big)$$ $$=\frac{1}{2}\sin^{-1}\Big(\frac{t-5}{9}\Big)$$ $$=\frac{1}{2}\sin^{-1}\Big(\frac{1}{9}(x^2-5)\Big)$$