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प्रश्न
The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]
विकल्प
- \[\frac{a^{x + \frac{1}{x}}}{\log_e a}\]
- \[\log_e a \cdot a^{x + \frac{1}{x}}\]
- \[\frac{a^{x + \frac{1}{x}}}{x} \log_e a\]
- \[x\frac{a^{x + \frac{1}{x}}}{\log_e a}\]
MCQ
उत्तर
\[\frac{a^{x + \frac{1}{x}}}{\log_e a}\]
\[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) \cdot a^{x + \frac{1}{x}} \]
\[ \therefore \int f\left( x \right)dx = \int\left( 1 - \frac{1}{x^2} \right) \cdot a^{x + \frac{1}{x}} dx\]
\[\text{Let }\left( x + \frac{1}{x} \right) = t\]
\[ \Rightarrow \left( 1 - \frac{1}{x^2} \right)dx = dt\]
\[ \therefore \int f\left( x \right)dx = \int a^t \cdot dt\]
\[ = \frac{a^t}{\log_e a} + C\]
\[ = \frac{a^{x + \frac{1}{x}}}{\log_e a} + C ...........\left( \because t = x + \frac{1}{x} \right)\]
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