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∫ 2 X + 1 ( X − 2 ) ( X − 3 ) D X - Mathematics

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प्रश्न

\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
बेरीज
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उत्तर

We have,
\[I = \int \frac{\left( 2x + 1 \right)dx}{\left( x - 2 \right) \left( x - 3 \right)}\]
\[\text{Let }\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} = \frac{A}{x - 2} + \frac{B}{x - 3}\]
\[ \Rightarrow \frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} = \frac{A\left( x - 3 \right) + B\left( x - 2 \right)}{\left( x - 2 \right) \left( x - 3 \right)}\]
\[ \Rightarrow 2x + 1 = A\left( x - 3 \right) + B\left( x - 2 \right)\]
\[\text{Putting }x - 3 = 0\]
\[ \Rightarrow x = 3\]
\[ \therefore 7 = A \times 0 + B \times \left( 3 - 2 \right)\]
\[ \Rightarrow B = 7\]
\[\text{Putting }x - 2 = 0\]
\[ \Rightarrow x = 2\]
\[ \therefore 5 = A\left( - 1 \right)\]
\[ \Rightarrow A = - 5\]
\[ \therefore I = - 5\int\frac{dx}{x - 2} + 7\int\frac{dx}{x - 3}\]
\[ = - 5 \log \left| x - 2 \right| + 7 \log \left| x - 3 \right| + C\]
\[ = \log \left| x - 3 \right|^7 - \log \left| x - 2 \right|^5 + C\]
\[ = \log \left| \frac{\left( x - 3 \right)^7}{\left( x - 2 \right)^5} \right| + C\]

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पाठ 19: Indefinite Integrals - Exercise 19.30 [पृष्ठ १७७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.30 | Q 52 | पृष्ठ १७७

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