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प्रश्न
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
योग
उत्तर
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)}dx\]
\[\text{Let x e}^x = t\]
\[ \Rightarrow \left( 1 \cdot e^x + \text{x e}^x \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( x + 1 \right) \text{e}^x dx = dt\]
\[Now, \int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)}dx\]
\[ = \int\frac{dt}{\sin^2 t}\]
\[ = \int {cosec}^2 \text{t dt} \]
\[ = - \text{cot} \left( t \right) + C\]
\[ = - \text{cot} \left( \text{x e}^x \right) + C\]
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