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We work out the following integral:
 $I\equiv \int _{0}^{1}{\frac {t^{4}(1t)^{4}}{1+t^{2}}}\,\mathrm {d} t\qquad \qquad \qquad (1)$
It is possible to divide polynomials in a manner that is analogous to long division of decimal numbers. By doing this it, can be shown that
 ${\frac {t^{4}(1t)^{4}}{1+t^{2}}}={\frac {t^{8}4t^{7}+6t^{6}4t^{5}+t^{4}}{1+t^{2}}}=t^{6}4t^{5}+5t^{4}4t^{2}+4{\frac {4}{1+t^{2}}}$
where −4 is the remainder of the polynomial division.
Using:
 $\int _{0}^{1}t^{n}\,\mathrm {d} t={\frac {1}{n+1}}$
for n=6, 5, 4, 2, and 0, respectively, one obtains
 $\int _{0}^{1}(t^{6}4t^{5}+5t^{4}4t^{2}+4)\,\mathrm {d} t={\frac {1}{7}}{\frac {4}{6}}+{\frac {5}{5}}{\frac {4}{3}}+{\frac {4}{1}}={\frac {22}{7}}$
The following holds
 $4\int _{0}^{1}{\frac {1}{1+t^{2}}}\,\mathrm {d} t=4\left[\arctan(t)\right]_{0}^{1}=4{\frac {\pi }{4}}=\pi$
The latter integral is easily evaluated by making the substitution
 $t=\tan(x)\,$
Hence
 $I={\frac {22}{7}}\pi$
The integrand (expression under the integral) of the integral in equation (1) is everywhere positive on the integration interval [0, 1] and, remembering that an integral can be defined as a sum of integrand values, it follows that $I\,$ is positive. Finally,
 ${\frac {22}{7}}\pi >0\quad \Longrightarrow \quad {\frac {22}{7}}>\pi$
which was to be proved.