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प्रश्न
Evaluate the following integral :-
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उत्तर
\[\text{Let }I = \int\frac{x^2 + x + 1}{\left( x^2 + 1 \right)\left( x + 2 \right)}dx\]
We express
\[\frac{x^2 + x + 1}{\left( x^2 + 1 \right)\left( x + 2 \right)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1}\]
\[ \Rightarrow x^2 + x + 1 = A\left( x^2 + 1 \right) + \left( Bx + C \right)\left( x + 2 \right)\]
\[\text{Equating the coefficients of x^2 , x and constants, we get}\]
\[1 = A + B and 1 = 2B + C and 1 = A + 2C\]
\[or A = \frac{3}{5} and B = \frac{2}{5} and C = \frac{1}{5} \]
\[ \therefore I = \int\left( \frac{\frac{3}{5}}{x + 2} + \frac{\frac{2}{5}x + \frac{1}{5}}{x^2 + 1} \right)dx\]
\[ = \frac{3}{5}\int\frac{1}{x + 2}dx + \frac{2}{5}\int\frac{x}{x^2 + 1} dx + \frac{1}{5}\int\frac{1}{x^2 + 1} dx\]
\[ = \frac{3}{5} I_1 + \frac{2}{5} I_2 + \frac{1}{5} I_3 ............(1)\]
\[\text{Now, }I_1 = \int\frac{1}{x + 2}dx\]
Let x + 2 = u
On differentiating both sides, we get
\[ dx = du\]
\[ \therefore I_1 = \int\frac{1}{u}du\]
\[ = \log\left| u \right| + c_1 \]
\[ = \log\left| x + 2 \right| + c_1 ............(2)\]
\[\text{And, }I_2 = \int\frac{x}{x^2 + 1} dx\]
\[\text{Let }\left( x^2 + 1 \right) = u\]
On differentiating both sides, we get
\[ 2x\ dx = du\]
\[ \therefore I_2 = \frac{1}{2}\int\frac{1}{u}du\]
\[ = \frac{1}{2}\log\left| u \right| + c_2 \]
\[ = \frac{1}{2}\log\left| x^2 + 1 \right| + c_2 ............(3)\]
\[\text{And, }I_3 = \int\frac{1}{x^2 + 1} dx\]
\[ = \tan^{- 1} x + c_3 ..............(4)\]
From (1), (2), (3) and (4), we get
\[ \therefore I = \frac{3}{5}\left( \log\left| x + 2 \right| + c_1 \right) + \frac{2}{5}\left( \frac{1}{2}\log\left| x^2 + 1 \right| + c_2 \right) + \frac{1}{5}\left( \tan^{- 1} x + c_3 \right)\]
\[ = \frac{3}{5}\log\left| x + 2 \right| + \frac{1}{5}\log\left| x^2 + 1 \right| + \frac{1}{5} \tan^{- 1} x + c\]
\[\text{Hence, }\int\frac{x^2 + x + 1}{\left( x^2 + 1 \right)\left( x + 2 \right)}dx = \frac{3}{5}\log\left| x + 2 \right| + \frac{1}{5}\log\left| x^2 + 1 \right| + \frac{1}{5} \tan^{- 1} x + c\]
