How can we evaluate $\int_{0}^{1}{1-x^{2n+1}\over (1-x^2)^{3/2}}\mathrm dx?$ Consider these two integrals

$$\int_{0}^{1}{1-x^{2n+1}\over (1-x^2)^{3/2}}\mathrm dx=I\tag1$$
$$\int_{0}^{1}{1-x^{2n}\over (1-x^2)^{3/2}}\mathrm dx={n\pi\over I}\tag2$$

An attempt: $(1)$
$x=\sin{u}$ then $dx=\cos{u}du$
$(1)$ becomes
$$\int_{0}^{\pi/2}(1-\sin^{2n+1}{u})\cdot{\mathrm du\over \cos^2{u}}\tag3$$
Applying integration by parts to $(3)$ seem long because of evaluating $\sin^{2n+1}{u}$
How else can we evaluate $(1)$ and $(2)$ using another way?
 A: Both $(1)$ and $(2)$ can be computed in terms of the $\Gamma$ function by exploiting integration by parts, the substitution $x=\sqrt{z}$ and Euler's Beta function. For instance
$$ \int_{0}^{1}\frac{1-x^{2n+1}}{(1-x^2)^{3/2}}\,dx = \frac{\sqrt{\pi}\,\,\Gamma(n+1)}{\Gamma\left(n+\frac{1}{2}\right)}\tag{1}$$
$$ \int_{0}^{1}\frac{1-x^{2n}}{(1-x^2)^{3/2}}\,dx = \frac{\sqrt{\pi}\,\,\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n\right)}\tag{2}$$
and the claim $(1)\cdot(2)=\pi\,n$ trivially follows.
A: Well after a long time I managed to come up with a formula for $I_{n}$
$I_{n} = \int_{0}^{\frac{\pi}{2}}\frac{1- \sin^{2n+1}u}{1-\sin^{2}u}du$
$I_{n} = -\int_{0}^{\frac{\pi}{2}}\frac{(\sin u-1)(\sin ^{2n}u + \sin ^{2n-1}u +... +\sin u + 1)}{(1-\sin u)(1+\sin u)}du$
$I_{n} =  \int_{0}^{\frac{\pi}{2}}\frac{(\sin u + 1)(\sin^{2n-1}u + \sin^{2n-3}u +...\sin u + 1)}{\sin u +1}du$
$I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}u + ... + \sin u du +\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin u + 1} du$
We can calculate $\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin u + 1} du = 1$
Then $I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}u + ... + \sin u du +1$
Consider $J_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}u du$
We can use reduction formula to derive that $J_{n} = \frac{2^{2n+1}n!(n+1)!}{(2n+2)!}$
Then $I_{n} = \sum_{i =0}^{n-1}  \frac{2^{2i+1}i!(i+1)!}{(2i+2)!} + 1$
Not sure how useful it is, but I'm pretty sure it's correct
