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प्रश्न
Evaluate the following integral:
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उत्तर
\[\text{Let }I = \int\frac{x^2}{x^4 + x^2 - 2}dx\]
We express
\[\frac{x^2}{x^4 + x^2 - 2} = \frac{x^2}{x^4 + 2 x^2 - x^2 - 2}\]
\[ = \frac{x^2}{\left( x^2 + 2 \right)\left( x^2 - 1 \right)}\]
\[ = \frac{A}{x^2 + 2} + \frac{B}{x^2 - 1}\]
\[ \Rightarrow x^2 = A\left( x^2 - 1 \right) + B\left( x^2 + 2 \right)\]
Equating the coefficients of `x^2` and constants, we get
\[1 = A + B\text{ and }0 = - A + 2B\]
\[\text{or }A = \frac{2}{3}\text{ and }B = \frac{1}{3}\]
\[ \therefore I = \int\left( \frac{\frac{2}{3}}{x^2 + 2} + \frac{\frac{1}{3}}{x^2 - 1} \right)dx\]
\[ = \frac{2}{3}\int\frac{1}{x^2 + 2}dx + \frac{1}{3}\int\frac{1}{x^2 - 1} dx\]
\[ = \frac{\sqrt{2}}{3} \tan^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{6}\log\left| \frac{x - 1}{x + 1} \right| + c\]
\[\text{Hence, }\int\frac{x^2}{x^4 + x^2 - 2}dx = \frac{\sqrt{2}}{3} \tan^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{6}\log\left| \frac{x - 1}{x + 1} \right| + c\]
