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Question
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Solution
\[Let\ I = \int_0^1 \frac{1 - x^2}{\left( 1 + x^2 \right)^2} d\ x . Then, \]
\[I = \int_0^1 \frac{\left( \frac{1}{x^2} - 1 \right)}{\left( x + \frac{1}{x} \right)^2} dx\]
\[Let\ x + \frac{1}{x} = t . Then, 1 - \frac{1}{x^2} dx = dt\]
\[When\ x = 0, t\ = \infty\ and\ x\ = 1, t = 2\]
\[ \therefore I = \int_\infty^2 \frac{- dt}{t^2}\]
\[ \Rightarrow I = \left[ \frac{1}{t} \right]_\infty^2 \]
\[ \Rightarrow I = \frac{1}{2} - 0\]
\[ \Rightarrow I = \frac{1}{2}\]
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