Question
Find:
$\int\frac{x^{2}}{x^{4} + x^{2} - 2}dx$
$\int\frac{x^{2}}{x^{4} + x^{2} - 2}dx$
✓
Answer
$\text{Let x}^{2} =\text{t} \therefore \frac{\text{x}^{2}}{\text{x}^{4} + \text{x}^{2} - 2} = \frac{\text{x}^{2}}{\text{(x}^{2} - 1)(\text{x}^{2} + 2)} = \frac{\text{t}}{(\text{t - 1)} (\text{t+2)}}=\frac{\text{A}}{\text{t - 1}} + \frac{\text{B}}{\text{t + 2}}$
Solving for A and B to get, $\text{A =}\frac{1}{3}, \text{B} = \frac{2}{3}$
$\int\frac{\text{x}^{2}}{\text{x}^{4} + \text{x}^{2} - 2}\text{dx} = \frac{1}{3}\int\frac{1}{\text{x}^{2} - 1} \text{dx} + \frac{2}{3}\int\frac{1}{\text{x}^{2} + 2}\text{dx} = \frac{1}{6}\log\bigg|\frac{\text{x - 1}}{\text{x + 1}}\bigg| + \frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}} + \text{C}$
Solving for A and B to get, $\text{A =}\frac{1}{3}, \text{B} = \frac{2}{3}$
$\int\frac{\text{x}^{2}}{\text{x}^{4} + \text{x}^{2} - 2}\text{dx} = \frac{1}{3}\int\frac{1}{\text{x}^{2} - 1} \text{dx} + \frac{2}{3}\int\frac{1}{\text{x}^{2} + 2}\text{dx} = \frac{1}{6}\log\bigg|\frac{\text{x - 1}}{\text{x + 1}}\bigg| + \frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}} + \text{C}$
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