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प्रश्न
Evaluate the following integral :-
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उत्तर
\[\text{Let }I = \int\frac{x}{\left( x^2 + 1 \right)\left( x - 1 \right)}dx\]
We express
\[\frac{x}{\left( x^2 + 1 \right)\left( x - 1 \right)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}\]
\[ \Rightarrow x = A\left( x^2 + 1 \right) + \left( Bx + C \right)\left( x - 1 \right)\]
Equating the coefficients of `x^2` , x and constants, we get
\[0 = A + B\text{ and }1 = - B + C\text{ and }0 = A - C\]
\[\text{or }A = \frac{1}{2}\text{ and }B = - \frac{1}{2}\text{ and }C = \frac{1}{2} \]
\[ \therefore I = \int\left( \frac{\frac{1}{2}}{x - 1} + \frac{- \frac{1}{2}x + \frac{1}{2}}{x^2 + 1} \right)dx\]
\[ = \frac{1}{2}\int\frac{1}{x - 1}dx - \frac{1}{2}\int\frac{x}{x^2 + 1} dx + \frac{1}{2}\int\frac{1}{x^2 + 1} dx\]
\[ = \frac{1}{2} I_1 - \frac{1}{2} I_2 + \frac{1}{2} I_3 ............(1)\]
\[\text{Now, }I_1 = \int\frac{1}{x - 1}dx\]
Let x - 1 = u
On differentiating both sides, we get
\[ dx = du\]
\[ \therefore I_1 = \int\frac{1}{u}du\]
\[ = \log\left| u \right| + c_1 \]
\[ = \log\left| x - 1 \right| + c_1 ..............(2)\]
\[\text{And, }I_2 = \int\frac{x}{x^2 + 1} dx\]
\[\text{Let }\left( x^2 + 1 \right) = u\]
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
\[ \therefore I = \frac{1}{2}\left( \log\left| x - 1 \right| + c_1 \right) - \frac{1}{2}\left( \frac{1}{2}\log\left| x^2 + 1 \right| + c_2 \right) + \frac{1}{2}\left( \tan^{- 1} x + c_3 \right)\]
\[ = \frac{1}{2}\log\left| x - 1 \right| - \frac{1}{4}\log\left| x^2 + 1 \right| + \frac{1}{2} \tan^{- 1} x + c\]
\[\text{Hence, }\int\frac{x}{\left( x^2 + 1 \right)\left( x - 1 \right)}dx = \frac{1}{2}\log\left| x - 1 \right| - \frac{1}{4}\log\left| x^2 + 1 \right| + \frac{1}{2} \tan^{- 1} x + c\]
