मराठी

2 ∫ 1 X + 3 X ( X + 2 ) D X

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प्रश्न

\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]

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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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पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 26 | पृष्ठ १२१

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