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प्रश्न
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
उत्तर
\[\text{ Let I} = \int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
\[\text{ Let } \tan^{- 1} x = t\]
\[ \Rightarrow \frac{dx}{1 + x^2} = dt\]
\[ \therefore I = \int e^t dt\]
\[ = e^t + C\]
\[ = e^{{tan}^{- 1}} x + C\]
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