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प्रश्न
उत्तर
We have,
\[I = \int\frac{\left( x^2 + 1 \right) dx}{\left( x - 2 \right)^2 \left( x + 3 \right)}\]
\[\text{Let }\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} = \frac{A}{x - 2} + \frac{B}{\left( x - 2 \right)^2} + \frac{C}{x + 3}\]
\[ \Rightarrow \frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} = \frac{A \left( x - 2 \right) \left( x + 3 \right) + B \left( x + 3 \right) + C \left( x - 2 \right)^2}{\left( x - 2 \right)^2 \left( x + 3 \right)}\]
\[ \Rightarrow x^2 + 1 = A \left( x^2 - 2x + 3x - 6 \right) + B \left( x + 3 \right) + C \left( x^2 - 4x + 4 \right)\]
\[ \Rightarrow x^2 + 1 = A \left( x^2 + x - 6 \right) + B \left( x + 3 \right) + C \left( x^2 - 4x + 4 \right)\]
Equating coefficients of like terms
\[A + C = 1 ..................(1)\]
\[A + B - 4C = 0 ...................(2)\]
\[ - 6A + 3B + 4C = 1 .....................(3)\]
Solving (1), (2) and (3), we get
\[A = \frac{3}{5}, B = 1\text{ and }C = \frac{2}{5}\]
\[ \therefore I = \frac{3}{5}\int\frac{dx}{x - 2} + \int\frac{dx}{\left( x - 2 \right)^2} + \frac{2}{5}\int\frac{dx}{x + 3}\]
\[ = \frac{3}{5} \log \left| x - 2 \right| + \left[ \frac{\left( x - 2 \right)^{- 2 + 1}}{- 2 + 1} \right] + \frac{2}{5} \log \left| x + 3 \right| + C\]
\[ = \frac{3}{5}\log \left| x - 2 \right| - \frac{1}{x - 2} + \frac{2}{5} \log \left| x + 3 \right| + C\]
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