Antiderivative of $\frac{(x+2)^2}{\sqrt{1-\frac{1}{x^2}}}$ I am struggling to find the general antiderivative of the following function $f$:
$$f(x)=\frac{(x+2)^2}{\sqrt{1-\frac{1}{x^2}}}$$
I am trying to find the improper integral from $1$ to $2$. I’m aware that the function is undefined at $1$ and I know the method to carry out.  I am just unsure on finding the general antiderivative of $f$. 
I tried substitution of $1/x$ but I was unable to proceed further. 
Can anyone help – I’ve spent a few days on this problem and don’t seem to be getting anywhere. Thanks.

$$\frac{(x+2)^2}{\sqrt{1-\dfrac1{x^2}}}=\frac{(x+4)(x^2-1)+5x+4}{\sqrt{x^2-1}}=x\sqrt{x^2-1}+4\sqrt{x^2-1}+\frac{5x}{\sqrt{x^2-1}}+\frac4{\sqrt{x^2-1}}.$$
Among these four terms, three have a quasi-immediate antiderivative,
$$I=\frac13(x^2-1)^{3/2}+4\int\sqrt{x^2-1}\,dx+5\sqrt{x^2-1}+4\text{ arcosh}(x).$$
Now by parts,
$$J=\int\sqrt{x^2-1}\,dx=x\sqrt{x^2-1}-\int\frac{x^2}{\sqrt{x^2-1}}dx=x\sqrt{x^2-1}-\int\frac{x^2-1+1}{\sqrt{x^2-1}}dx
\\=x\sqrt{x^2-1}-\text{arcosh}(x)-J.$$
