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प्रश्न
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उत्तर
We have,
\[I = \int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)}dx\]
\[\text{Let }\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 4}\]
\[ \Rightarrow \frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} = \frac{A \left( x^2 + 4 \right) + \left( Bx + C \right) \left( x + 2 \right)}{\left( x + 2 \right) \left( x^2 + 4 \right)}\]
\[ \Rightarrow 18 = A x^2 + 4A + B x^2 + 2Bx + Cx + 2C\]
\[ \Rightarrow 18 = \left( A + B \right) x^2 + x \left( 2B + C \right) + 4A + 2C\]
\[\text{Equating coefficients of like terms}\]
\[A + B = 0 . . . . . \left( 1 \right)\]
\[2B + C = 0 . . . . . \left( 2 \right)\]
\[4A + 2C = 18 . . . . . \left( 3 \right)\]
\[\text{Solving (1), (2) and (3), we get}\]
\[A = \frac{9}{4}\]
\[B = - \frac{9}{4}\]
\[C = \frac{9}{2}\]
\[ \therefore \frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} = \frac{9}{4 \left( x + 2 \right)} + \frac{- \frac{9}{4}x + \frac{9}{2}}{x^2 + 4}\]
\[ \Rightarrow \frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} = \frac{9}{4 \left( x + 2 \right)} - \frac{9}{4} \left( \frac{x}{x^2 + 4} \right) + \frac{9}{2 \left( x^2 + 4 \right)}\]
\[ \Rightarrow \int\frac{18 dx}{\left( x + 2 \right) \left( x^2 + 4 \right)} = \frac{9}{4}\int\frac{dx}{x + 2} - \frac{9}{4}\int\frac{x dx}{x^2 + 4} + \frac{9}{2}\int\frac{dx}{x^2 + 2^2}\]
\[\text{Let }x^2 + 4 = t\]
\[ \Rightarrow 2xdx = dt\]
\[ \Rightarrow x dx = \frac{dt}{2}\]
\[ \therefore I = \frac{9}{4}\int\frac{dx}{x + 2} - \frac{9}{8}\int\frac{dt}{t} + \frac{9}{2}\int\frac{dx}{x^2 + 2^2}\]
\[ = \frac{9}{4} \log \left| x + 2 \right| - \frac{9}{8} \log \left| t \right| + \frac{9}{2} \times \frac{1}{2} \tan^{- 1} \left( \frac{x}{2} \right) + C'\]
\[ = \frac{9}{4} \log \left| x + 2 \right| - \frac{9}{8} \log \left| x^2 + 4 \right| + \frac{9}{4} \tan^{- 1} \left( \frac{x}{2} \right) + C'\]
