Question with Solution

$1={\int }_0^1{\left(1-{\mathrm{x}}_1^4\right)}^7-\underset{\pi }{1\mathrm{dx}}$

$$\begin{array}{rlrlr}& ={\left[\mathrm{x}{\left(1-{\mathrm{x}}^4\right)}^7\right]}_0^1+7\times 4{\int }_0^1\mathrm{x}{\left(1-{\mathrm{x}}^4\right)}^6{\mathrm{x}}^3\mathrm{dx}& =-28{\int }_0^1{\left(1-{\mathrm{x}}^4\right)}^6\mathrm{dx}+28{\int }_0^1{\left(1-{\mathrm{x}}^4\right)}^6\mathrm{dx}=-28\mathrm{I}+28{\int }_0^1{\left(1-{\mathrm{x}}^4\right)}^6\mathrm{dx}& 29\mathrm{I}-28{\int }_0^1\left(1-{\mathrm{x}}^4\right)\mathrm{dx}& \frac{29{\int }_0^1{\left(1-{\mathrm{x}}^4\right)}^7\mathrm{dx}}{4{\int }_0^1{\left(1-{\mathrm{x}}^4\right)}^7\mathrm{dx}}=7\end{array}$$