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Question
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Solution
\[Let\ x^2 = t . Then, 2x\ dx = dt\]
\[When\ x = 1, t = 1\ and\ x = 2, t = 4\]
\[ \therefore I = \int_1^2 \frac{3x}{9 x^2 - 1} d x\]
\[ \Rightarrow I = \frac{3}{2} \int_1^4 \frac{dt}{9t - 1}\]
\[ \Rightarrow I = \frac{3}{18} \left[ \log \left( 9t - 1 \right) \right]_1^4 \]
\[ \Rightarrow I = \frac{3}{18}\left( \log 35 - \log 8 \right)\]
\[ \Rightarrow I = \frac{\left( \log 35 - \log 8 \right)}{6}\]
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