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प्रश्न
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
उत्तर
\[\int_1^2 \frac{x + 3}{x\left( x + 2 \right)} d x\]
\[ = \int_1^2 \frac{x + 2 + 1}{x\left( x + 2 \right)} d x\]
\[ = \int_1^2 \frac{1}{x}dx + \int_1^2 \frac{1}{x\left( x + 2 \right)}dx\]
\[ = \int_1^2 \frac{1}{x}dx + \frac{1}{2} \int_1^2 \frac{\left( x + 2 \right) - x}{x\left( x + 2 \right)}dx\]
\[ = \int_1^2 \frac{1}{x}dx + \frac{1}{2} \int_1^2 \frac{1}{x}dx - \frac{1}{2} \int_1^2 \frac{1}{x + 2}dx\]
\[ = \frac{3}{2} \int_1^2 \frac{1}{x}dx - \frac{1}{2} \int_1^2 \frac{1}{x + 2}dx\]
\[ = \frac{3}{2} \left[ \log x \right]_1^2 - \frac{1}{2} \left[ \log\left( x + 2 \right) \right]_1^2 \]
\[ = \frac{3}{2}\log2 - \frac{1}{2}\log4 + \frac{1}{2}\log3\]
\[ = \frac{3}{2}\log2 - \log2 + \frac{1}{2}\log3\]
\[ = \frac{1}{2}\log2 + \frac{1}{2}\log3\]
\[ = \frac{1}{2}\log6\]
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