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प्रश्न
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उत्तर
\[\int\frac{\left( x^2 + x + 5 \right)}{\left( 3x + 2 \right)}dx\]
\[ = \frac{1}{9}\int\frac{9 x^2 + 9x + 45}{\left( 3x + 2 \right)}dx\]
\[ = \frac{1}{9}\left[ \int\frac{9 x^2 - 4}{3x + 2}dx + \int\frac{9x + 6}{3x + 2}dx + \int\frac{43}{3x + 2}dx \right]\]
\[ = \frac{1}{9}\left[ \int\frac{\left( 3x - 2 \right)\left( 3x + 2 \right)}{\left( 3x + 2 \right)}dx + \int\frac{3\left( 3x + 2 \right)}{3x + 2}dx + 43\int\frac{dx}{3x + 2} \right]\]
\[ = \frac{1}{9}\left[ \int\left( 3x - 2 \right) dx + 3\int1dx + 43\int\frac{dx}{3x + 2} \right]\]
\[ = \frac{1}{9}\left[ \left( 3\frac{x^2}{2} - 2x \right) + 3x + \frac{43}{3} \text{ln}\left| 3x + 2 \right| + C \right]\]
\[ = \frac{1}{9}\left[ \frac{3}{2} x^2 + x - \frac{43}{3} \text{ln }\left| 3x + 2 \right| + C \right]\]
\[ = \frac{1}{6} x^2 + \frac{1}{9}x - \frac{43}{27} \text{ln }\left| 3x + 2 \right| + C\]
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