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प्रश्न
उत्तर
We have,
\[I = \int\frac{dx}{x \log x\left( 2 + \log x \right)}\]
Putting log x = t
\[ \Rightarrow \frac{1}{x} dx = dt\]
\[ \therefore I = \int\frac{dt}{t \left( t + 2 \right)}\]
\[\text{Let }\frac{1}{t \left( t + 2 \right)} = \frac{A}{t} + \frac{B}{t + 2}\]
\[ \Rightarrow \frac{1}{t \left( t + 2 \right)} = \frac{A\left( t + 2 \right) + Bt}{t \left( t + 2 \right)}\]
\[ \Rightarrow 1 = A \left( t + 2 \right) + Bt\]
Putting t + 2 = 0
\[ \Rightarrow t = - 2\]
\[1 = A \times 0 + B \left( - 2 \right)\]
\[ \Rightarrow B = - \frac{1}{2}\]
Putting t = 0
\[1 = A \left( 0 + 2 \right) + B \times 0\]
\[ \Rightarrow A = \frac{1}{2}\]
Then,
\[I = \frac{1}{2}\int\frac{dt}{t} - \frac{1}{2}\int\frac{dt}{t + 2}\]
\[ = \frac{1}{2} \left[ \log \left| t \right| - \log \left| t + 2 \right| \right] + C\]
\[ = \frac{1}{2} \log \left| \frac{t}{t + 2} \right| + C\]
\[ = \frac{1}{2} \log \left| \frac{\log x}{\log x + 2} \right| + C\]
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