Determining $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$ 
$$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$$

Attempt:
Simplification of the root factor:
$$\sqrt{x^4+2x^3-x^2+2x+1}=\frac{1}{x}\sqrt{\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)-1}.$$
Arranging the rest of the factors as:
$$\frac{x-1}{x^2(x+1)}=\frac{x^2-1}{x^2(x+1)^2}$$
Now I did the following substitution: let $(x+\frac{1}{x})=t$, so $(1-\frac{1}{x^2})dx=dt$
Arranging the integral: $$I=\int \frac{\sqrt{t^2+2t-3}}{t+2}dt.$$
But what to do next?
I tried integration by parts for this but couldn't simplify my result.
 A: Maybe it's not the most rapid way, but it seems work. We have, taking $u=t+1
 $ $$\int\frac{\sqrt{t^{2}+2t-3}}{t+2}dt=\int\frac{\sqrt{u^{2}-4}}{u+1}du
 $$ and taking $u=2\sec\left(v\right)
 $ we have $$=4\int\frac{\tan^{2}\left(v\right)\sec\left(v\right)}{2\sec\left(v\right)+1}dv=4\int\frac{\tan^{2}\left(v\right)}{\cos\left(v\right)+2}dv
 $$ using $\tan^{2}\left(v\right)=\sec^{2}\left(v\right)+1
 $. Now we can take $w=\tan\left(v/2\right)
 $ and get $$=32\int\frac{w^{2}}{w^{6}+w^{4}-5w^{2}+3}dw=32\int\frac{w^{2}}{\left(w^{2}-1\right)^{2}\left(w^{2}+3\right)}dw
 $$ and now using a boring partial fractions you can transform the integral in a tractable form.
A: $\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{(x-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx = \int\frac{(x^2-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x^2+2x+1)}dx$$
Above we multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $(x+1).$
$$\displaystyle = \int\frac{\left(1-\frac {1}{x^2}\right)\cdot \sqrt{x^2\cdot \left(x^2+2x-1+\frac{2}{x}+\frac{1}{x^2}\right)}}{ \left(x+2+\frac{1}{x}\right)}dx$$
Now Let $ \displaystyle \left(x+\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1-\frac{1}{x^2}\right)dx = dt$
So Integral $$\displaystyle = \int\frac{\sqrt{t^2+2t-3}}{t+2}dt  = \frac{t^2+2t-3}{(t+2)\sqrt{t^2+2t-3}}dt = \int\frac{t(t+2)-3}{(t+2)\sqrt{t^2+2t-3}}dt$$
So Integral $$\displaystyle = \underbrace{\int\frac{t}{\sqrt{t^2+2t-3}}dt}_{I} - \underbrace{\int\frac{3}{(t+2)}\cdot \frac{1}{\sqrt{t^2+2t-3}}dt}_{J}..........\color{\red}\checkmark.$$
So $$\displaystyle I = \int\frac{t}{\sqrt{t^2+2t-3}}dt = \int\frac{(t+1)-1}{\sqrt{(t-1)^2-2^2}} = \int\frac{(t-1)}{\sqrt{(t-1)^2-2^2}}-\int\frac{1}{\sqrt{(t-1)^2-2^2}}dt$$
Now Let $(t-1) = z\;\;,$ Then $dt = dz$
So $$\displaystyle I = \int\frac{z}{\sqrt{z^2-2^2}}dz-\int\frac{1}{\sqrt{z^2-2^2}}dz = \sqrt{z^2-4}-\ln \left|(t+1)+\sqrt{t^2+2t-3}\right|$$
Now $$\displaystyle J = 3\int\frac{1}{(t+2)\sqrt{t^2+2t-3}}dt = 3\int\frac{1}{(t+2)\sqrt{(t+2)^2-2(t+2)+1-4}}$$
Now Let $(t+2) = u\;,$ Then $dt = du$ and Integral $$\displaystyle = 3\int\frac{1}{u\sqrt{u^2-2u+1-4}}=3\int\frac{1}{u\sqrt{(u-1)^2-4}}du$$
Now $\displaystyle (u-1) = 2\sec \theta \;, $ Then $du= 2\sec \theta \cdot \tan \theta.$
Now after that You can Solve It. 
