Let $I_1 =\displaystyle\int^1_0(1-x^4)^7\, \mathrm d x$ and $I_2 =\displaystyle\int^1_0(1-x^4)^6\, \mathrm d x$

Integrate the numerator using integration by parts:
\begin{align}
\require{cancel} I_1 =\int^1_01\cdot(1-x^4)^7 \mathrm d x&= \cancelto{0}{x(1-x^4)^7\big]^1_0} -\int^1_0x\cdot(-4x^3)\cdot7(1-x^4)^6\,\mathrm d x \\
&= 28\int^1_0x^4(1-x^4)^6\, \mathrm d x\\
&= -28\int^1_0(\color{red}{1-x^4}-1)(1-x^4)^6\,\mathrm d x \\
I_1&= -28\underset{I_1}{\underbrace{\int^1_0\color{red}{(1-x^4)}(1-x^4)^6\,\mathrm d x}}+28\underset{I_2}{\underbrace{\int^1_0(1-x^4)^6\,\mathrm d x}} \\