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प्रश्न
\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]
बेरीज
उत्तर
\[\int 5^{x + \tan^{- 1} x} \cdot \left( \frac{x^2 + 2}{x^2 + 1} \right)dx\]
\[\text{Let x} + \tan^{- 1} x = t\]
\[\left( 1 + \frac{1}{1 + x^2} \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( \frac{x^2 + 1 + 1}{x^2 + 1} \right)dx = dt\]
\[ \Rightarrow \left( \frac{x^2 + 2}{x^2 + 1} \right)dx = dt\]
\[Now, \int 5^{x + \tan^{- 1} x} \cdot \left( \frac{x^2 + 2}{x^2 + 1} \right)dx\]
\[ = \int 5^t dt\]
\[ = \frac{5^t}{\log 5} + C\]
\[ = \frac{5^{x + \tan^1 x}}{\log 5} + C\]
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