∫0∞ x2dxx2+a2x2+b2=?

∫0∞ x2dxx2+a2x2+b2=∫0∞ x2+a2−a2x2+a2x2+b2dx∫0∞ 1x2+b2dx−a2∫0∞ 1x2+a2x2+b2dx=1btan−1⁡xb1∞−a2a2−b2∫0∞ 1x2+b2−1x2+a2dx=1b⋅π2−a2a2−b21btan−1⁡xb−1atan−1⁡xa⌋∞=π2(a+b).

∫0∞ x2dxx2+a2x2+b2=?

∫0∞ x2dxx2+a2x2+b2=∫0∞ x2+a2−a2x2+a2x2+b2dx∫0∞ 1x2+b2dx−a2∫0∞ 1x2+a2x2+b2dx=1btan−1⁡xb1∞−a2a2−b2∫0∞ 1x2+b2−1x2+a2dx=1b⋅π2−a2a2−b21btan−1⁡xb−1atan−1⁡xa⌋∞=π2(a+b).

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$\int_{0}^{\infty} \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = ?$

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$\frac{π}{2 (a - b)}$

$\frac{π}{2 (b - a)}$

$\frac{π}{(a + b)}$

$\frac{π}{2 (a + b)}$

answer is D.

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Detailed Solution

$\begin{array}{l} \int_{0}^{\infty} \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \int_{0}^{\infty} \frac{(x^{2} + a^{2}) - a^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x \\ \int_{0}^{\infty} \frac{1}{x^{2} + b^{2}} d x - a^{2} \int_{0}^{\infty} \frac{1}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x \\ = [{\frac{1}{b} \tan^{- 1} \frac{x}{b}|}_{1}^{\infty} - \frac{a^{2}}{(a^{2} - b^{2})} \int_{0}^{\infty} (\frac{1}{x^{2} + b^{2}} - \frac{1}{x^{2} + a^{2}}) d x \\ = \frac{1}{b} \cdot \frac{π}{2} - \frac{a^{2}}{(a^{2} - b^{2})} [\frac{1}{b} \tan^{- 1} \frac{x}{b} - {\frac{1}{a} \tan^{- 1} \frac{x}{a}|}_{⌋}^{\infty} = \frac{π}{2 (a + b)} . \end{array}$