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प्रश्न
\[\int\frac{\cos 2x + x + 1}{x^2 + \sin 2x + 2x} dx\]
योग
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उत्तर
\[\text{Let I }= \int\frac{\cos2x + x + 1}{x^2 + \sin2x + 2x}dx\]
\[Putting\ x^2 + \sin2x + 2x = t\]
\[ \Rightarrow 2x + 2\cos 2x + 2 = \frac{dt}{dx}\]
\[ \Rightarrow \left( x + \cos 2x + 1 \right)dx = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{1}{t}dt\]
\[ = \frac{1}{2}\text{ln}\left| t \right| + C\]
\[ = \frac{1}{2} \text{ln }\left| x^2 + \sin2x + 2x \right| + C \left[ \because t = x^2 + \sin 2x + 2x \right]\]
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