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Properties of Inequalities

∫ x x 2 -a2 x...

Question

$\int \frac{x}{(x^{2} - a^{2}) (x^{2} - b^{2})} d x$

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Solution

$We have, I = \int \frac{x d x}{(x^{2} - a^{2}) (x^{2} - b^{2})} Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{(t - a^{2}) (t - b^{2})} Let \frac{1}{(t - a^{2}) (t - b^{2})} = \frac{A}{t - a^{2}} + \frac{B}{t - b^{2}} \Rightarrow \frac{1}{(t - a^{2}) (t - b^{2})} = \frac{A (t - b^{2}) + B (t - a^{2})}{(t - a^{2}) (t - b^{2})} \Rightarrow 1 = A (t - b^{2}) + B (t - a^{2}) Putting t = b^{2} 1 = A \times 0 + B (b^{2} - a^{2}) \Rightarrow B = \frac{1}{b^{2} - a^{2}} Putting t = a^{2} 1 = A (a^{2} - b^{2}) + B \times 0 \Rightarrow A = \frac{1}{a^{2} - b^{2}} I = \frac{1}{2} \int \frac{d t}{(t - a^{2}) (t - b^{2})} = \frac{1}{2 (a^{2} - b^{2})} \int \frac{d t}{t - a^{2}} + \frac{1}{2 (b^{2} - a^{2})} \int \frac{d t}{t - b^{2}} = \frac{1}{2 (a^{2} - b^{2})} \log |t - a^{2}| + \frac{1}{2 (b^{2} - a^{2})} \log |t - b^{2}| + C = \frac{1}{2 (a^{2} - b^{2})} [\log |t - a^{2}| - \log |t - b^{2}|] + C = \frac{1}{2 (a^{2} - b^{2})} [\log |\frac{t - a^{2}}{t - b^{2}}|] + C = \frac{1}{2 (a^{2} - b^{2})} \log |\frac{x^{2} - a^{2}}{x^{2} - b^{2}}| + C$

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