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प्रश्न
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
उत्तर
\[\int\frac{x + \cos6x}{3 x^2 + \sin6x}dx\]
\[ \text{ Let }\left( 3 x^2 + \sin6x \right) = t\]
\[ \text{On differentiating both sides, we get}\]
\[ \left( 6x + 6\cos6x \right) dx = dt\]
\[ \therefore \int\frac{x + \cos6x}{3 x^2 + \sin6x}dx = \frac{1}{6}\int\frac{1}{t}dt\]
\[ = \frac{1}{6}\text{ log}\left| t \right| + c\]
\[ = \frac{1}{6}\text{ log}\left| 3 x^2 + \sin6x \right| + c\]
\[\text{ Hence,} \int\frac{x + \cos6x}{3 x^2 + \sin6x}dx = \frac{1}{6}\text{ log}\left| 3 x^2 + \sin6x \right| + c\]
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