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प्रश्न
उत्तर
We have,
\[I = \int\frac{\left( 2 x^2 + 7x - 3 \right) dx}{x^2 \left( 2x + 1 \right)}\]
\[\text{Let }\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x + 1}\]
\[ \Rightarrow \frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} = \frac{A \left( x \right) \left( 2x + 1 \right) + B \left( 2x + 1 \right) + C x^2}{x^2 \left( 2x + 1 \right)}\]
\[ \Rightarrow 2 x^2 + 7x - 3 = A \left( 2 x^2 + x \right) + B \left( 2x + 1 \right) + C x^2 \]
\[ \Rightarrow 2 x^2 + 7x - 3 = \left( 2A + C \right) x^2 + \left( A + 2B \right)x + B\]
\[\text{Equating coefficients of like terms}\]
\[2A + C = 2 . . . . . \left( 1 \right)\]
\[A + 2B = 7 . . . . . \left( 2 \right)\]
\[B = - 3 . . . . . \left( 3 \right)\]
\[\text{Solving (1), (2) and (3), we get}\]
\[A = 13\]
\[B = - 3\]
\[C = - 24\]
\[ \therefore \frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} = \frac{13}{x} - \frac{3}{x^2} - \frac{24}{2x + 1}\]
\[ \Rightarrow I = 13\int\frac{dx}{x} - 3\int x^{- 2} dx - 24\int\frac{dx}{2x + 1}\]
\[ = 13 \log \left| x \right| + \frac{3}{x} - 24 \frac{\log \left| 2x + 1 \right|}{2} + C\]
\[ = 13 \log \left| x \right| + \frac{3}{x} - 12 \log \left| 2x + 1 \right| + C\]
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