Calculation of $$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$$
Try: put $x=t^2$ and $dx = 2tdt$
Let $$I = 2\int\frac{(t^2-1)\sqrt{t^8+2t^6-t^4+2t^2+1}}{t(t^2+1)}dt$$
$$I = 2\int \frac{\bigg(1-\frac{1}{t^2}\bigg)\sqrt{t^8+2t^6-t^4+2t^2+1}}{(t+\frac{1}{t})}dt$$
could some help me to solve it , Thanks
Note first that $x^4+2x^3-x^2+2x+1$ has symmetric coefficients, so it can be expressed in terms of $(x+1/x)$. In particular, $$ \begin{align} x^4+2x^3-x^2+2x+1 &= x^2\left(\left(x+\frac 1x\right)^2+2\left(x+\frac 1x\right)-3\right) \end{align} $$ Bearing this in mind, we use $u = x + \frac 1x$, then $du = 1 - \frac{1}{x^2} dx$. Now, \begin{align} \int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}\,dx&=\int \frac{(x-1)\sqrt{u^2+2u-3}}{x(x+1)}\,dx\\ &=\int \frac{(x^2-1)\sqrt{u^2+2u-3}}{x(x+1)^2}\,dx\\ &=\int \frac{x(1-\frac{1}{x^2})\sqrt{u^2+2u-3}}{(x+1)^2}\,dx\\ &=\int \frac{x\sqrt{u^2+2u-3}}{(x+1)^2}\,du\\ &=\int \frac{\sqrt{u^2+2u-3}}{x+2+\frac 1x}\,du\\ &=\int \frac{\sqrt{u^2+2u-3}}{u+2}\,du. \end{align} This is now a much more routine integral (Wolfram Alpha manages), which I leave to you.
