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प्रश्न
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उत्तर
\[\text{Let I} = \int\frac{10 x^9 + {10}^x \log_e 10}{{10}^x + x^{10}}dx\]
\[\text{Putting }{10}^x + x^{10} = t\]
\[ \Rightarrow {10}^x \log_e 10 + 10 x^9 = \frac{dt}{dx}\]
\[ \Rightarrow \left( {10}^x \log_e 10 + 10 x^9 \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln} \left| t \right| + C\]
\[ = \text{ln} \left| {10}^x + x^{10} \right| + C \left[ \because t = {10}^x + x^{10} \right]\]
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