Nice Exercise shared in Art of mathematics page

Exercise:

Evaluate, $\int_{0}^{1}{\frac{12dx\left( {{\left( dx \right)}^{2}}-1 \right)}{\left( 1+\frac{1}{dx} \right)\left( dx-1 \right)\left( d\left( {{x}^{3}}+3x \right) \right)}=??}$

Solution: we have \[d\left( {{x}^{3}}+3x \right)=\left( 3{{x}^{2}}+3 \right)dx\] and\[{{\left( dx \right)}^{2}}-1=\left( dx-1 \right)\left( dx+1 \right)\] then

$\int_{0}^{1}{\frac{12dx\left( {{\left( dx \right)}^{2}}-1 \right)}{\left( 1+\frac{1}{dx} \right)\left( dx-1 \right)\left( d\left( {{x}^{3}}+3x \right) \right)}=\int_{0}^{1}{\frac{12dx\left( dx+1 \right)}{\left( 1+\frac{1}{dx} \right)\left( 3{{x}^{2}}+3 \right)dx}}}$

\[=\int_{0}^{1}{\frac{12dx}{\frac{\left( 3{{x}^{2}}+3 \right)dx}{dx}}=\int_{0}^{1}{\frac{12}{3{{x}^{2}}+3}dx}=4\int_{0}^{1}{\frac{dx}{{{x}^{2}}+1}}=4\left( \arctan x \right)_{0}^{1}=\pi }\]