Definite integral of $\int_0^{1}x^{7}\sqrt{\frac{1+x^{2}}{1-x^{2}}}dx$. When i integrate $\int_{0}^{1}x^{7}\sqrt{\frac{1+x^{2}}{1-x^{2}}}dx$, putting 
$t^{2}=1-x^{2}$ then integration $\int{(1-t^{2})}^{3}\sqrt{t^{2}+2}dt$, but i don't understand how to proceed . please someone help me. Thank you.
 A: One way to proceed further is an Euler substitution,
$$u=\sqrt{t^2+2}-t$$ this will reduce the integral to a rational function.
Another way to proceed from the beginning is to set $y=x^2$ then you get
$$\frac{1}{2}\int y^3\sqrt{\frac{1+y}{1-y}}dy$$
and now the substitution $$z=\sqrt{\frac{1+y}{1-y}}$$ will again give a rational expression, this method is probably neater.
A: Substitute $\text{u}=\frac{1}{1-x^2}$:
$$\mathcal{I}=\int x^7\cdot\sqrt{\frac{1+x^2}{1-x^2}}\space\text{d}x=\frac{1}{2}\int\frac{\left(\text{u}-1\right)^3\cdot\sqrt{2\text{u}-1}}{\text{u}^5}\space\text{d}\text{u}$$
Now, substitute $\text{v}=\sqrt{2\text{u}-1}$:
$$\mathcal{I}=2\int\frac{\text{v}^2\cdot\left(\text{v}^6-3\text{v}^4+3\text{v}^2-1\right)}{\left(1+\text{v}^2\right)^5}\space\text{d}\text{v}$$
And now you do partial fraction decomposition:
$$\frac{\text{v}^2\cdot\left(\text{v}^6-3\text{v}^4+3\text{v}^2-1\right)}{\left(1+\text{v}^2\right)^5}=\frac{1}{1+\text{v}^2}-\frac{7}{\left(1+\text{v}^2\right)^2}+\frac{18}{\left(1+\text{v}^2\right)^3}-\frac{20}{\left(1+\text{v}^2\right)^4}+\frac{8}{\left(1+\text{v}^2\right)^5}$$
A: HINT:
For $$\int x^{2n+1}f(x^2)dx$$ I'll start with $x^2=u$
let $\sqrt{\dfrac{1+x^2}{1-x^2}}=u\implies x^2=\dfrac{u^2-1}{u^2+1}=1-\dfrac2{1+u^2}$
