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Question
Solution
\[\text{We have}, \]
\[I = \int_0^1 \frac{2x}{1 + x^2} d x\]
\[\text{Putting} 1 + x^2 = t\]
\[ \Rightarrow 2x\ dx = dt\]
\[\text{When } x \to 0; t \to 1\]
\[\text{And } x \to 1; t \to 2\]
\[ \therefore I = \int_1^2 \frac{d t}{t}\]
\[ = \left[ \log_e \left| t \right| \right]_1^2 \]
\[ = \log_e 2 - \log_e 1\]
\[ = \log_e 2 - 0\]
\[ = \log_e 2\]
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