How can I prove this reduction formula for $\int^1_0{x(1-2x^4)^n}dx$ The exercise in my textbook states

You are given that $$I_n=\int^1_0{x(1-2x^4)^n}dx$$  
Show that $$I_n=\frac{(-1)^n}{4n +2} + \frac{2n}{2n+1}I_{n-1}$$

I have started out by splitting the power
$$\int^1_0{x(1-2x^4)^n}dx=\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx$$
$$\int^1_0{x(1-2x^4)^{n-1}}dx + \int^1_0{-2x^4(1-2x^4)^{n-1}}dx$$
$$I_n=I_{n-1} + \int^1_0{-2x^5(1-2x^4)^{n-1}}dx$$
Then setting up the second integral for integration by parts
$$ I_n=I_{n-1} + \frac{1}{2}\int^1_0{x^2*-4x^3(1-2x^4)^{n-1}}dx $$
Differentiating $x^2$ and integrating $-4x^3(1-2x^4)^{n-1}$
$$I_n=I_{n-1} + \frac{1}{2}\left(\left[\frac{x^2(1-2x^4)}{n}\right]^1_0 -\frac{2}{n}\int^1_0{x(1-2x^4)^n}dx\right)$$
Evaluating $\left[\frac{x^2(1-2x^4)}{n}\right]^1_0$
$$I_n=I_{n-1} + \frac{1}{2}\left(\frac{(-1)^n}{n} -\frac{2}{n}I_n \right)$$
Collecting like terms
$$I_n=I_{n-1} +\frac{(-1)^n}{2n} -\frac{1}{n}I_n$$
$$I_n=\frac{1}{1+\frac{1}{n}}I_{n-1} + \frac{(-1)^n}{2n(1+\frac{1}{n})}$$
Simplifying the fractions
$$I_n=\frac{n}{1+n}I_{n-1} + \frac{(-1)^n}{2n+2}$$
However this is not equivelant to the expression in the question, have I messed up somewhere or missed something out?

$$I_n=\int_0^1x(1-2x^4)^ndx$$
$u=x^2\Rightarrow dx=\frac{du}{2x}$ and so:
$$I_n=\frac12\int_0^1(1-2u^2)^ndu$$

$$2I_n=\int_0^1(1+\sqrt{2}u)^n(1-\sqrt{2}u)^ndu$$
If we do IBP:
$$a'=(1+\sqrt{2}u)^n\Rightarrow a=\frac{1}{\sqrt{2}(n+1)}(1+\sqrt{2}u)^{n+1}$$
$$b=(1-\sqrt{2}u)^n\Rightarrow b'=-\sqrt{2}n(1-\sqrt{2}u)^{n-1}$$
so:
$$2I_n=\frac{1}{\sqrt{2}(n+1)}\left[(1-2u^2)^n(1+\sqrt{2}u)\right]_0^1+\frac{n}{n+1}\int_0^1(1-2u^2)^{n-1}(1+\sqrt{2}u)^2du$$
$$2I_n=\frac{(-1)^n(1+\sqrt{2})-1}{\sqrt{2}(n+1)}+\frac{n}{n+1}\left(I_{n-1}+2\sqrt{2}\int_0^1u(1-2n^2)^{n-1}du+2\int_0^1u^2(1-2u^2)^{n-1}du\right)$$
