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प्रश्न
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उत्तर
\[Let\ x^2 = t\ . Then, 2x\ dx\ = dt\]
\[When\ x = 2, t = 4 and\ x\ = 4, t = 16 . \]
\[ \therefore I = \int_2^4 \frac{x}{x^2 + 1} d x\]
\[ \Rightarrow I = \int_4^{16} \frac{1}{2}\frac{dt}{t + 1}\]
\[ \Rightarrow I = \frac{1}{2} \left[ \log \left( t + 1 \right) \right]_4^{16} \]
\[ \Rightarrow I = \frac{1}{2} \log 17 - \frac{1}{2} \log 5\]
\[ \Rightarrow I = \frac{1}{2} \log \frac{17}{5}\]
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