For a≥2, if the value of the definite integral ∫0∞dxa2+x−1x2 equals π10 ,then the value of a is

I=∫0∞dxx2+1x2+(a2−2)=∫0∞x2dxx4+(a2−2)x2+1, Put a2−2=k =∫0∞x2dxx4+kx2+1=12∫0∞(x2+1)+(x2−1)x4+kx2+1dx =12∫0∞1+(1x2)x2+(1x2)+k⏟I1dx+12∫0∞1−(1x2)x2+(1x2)+k⏟I2dxNow proceed, I1=π2a and I2=0,∴ I=π2a=π10⇒ a=5

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For $a \geq 2,$ if the value of the definite integral $\int_{0}^{\infty} \frac{d x}{a^{2} + {(x - (\frac{1}{x}))}^{2}}$ equals $\frac{π}{10}$ ,then the value of a is

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answer is 5.

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Detailed Solution

$I = \int_{0}^{\infty} \frac{d x}{x^{2} + \frac{1}{x^{2}} + (a^{2} - 2)} = \int_{0}^{\infty} \frac{x^{2} d x}{x^{4} + (a^{2} - 2) x^{2} + 1}, P u t a^{2} - 2 = k = \int_{0}^{\infty} \frac{x^{2} d x}{x^{4} + k x^{2} + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + 1) + (x^{2} - 1)}{x^{4} + k x^{2} + 1} d x = \frac{1}{2} \underset{I_{1}}{\underset{⏟}{\int_{0}^{\infty} \frac{1 + (\frac{1}{x^{2}})}{x^{2} + (\frac{1}{x^{2}}) + k}}} d x + \frac{1}{2} \underset{I_{2}}{\underset{⏟}{\int_{0}^{\infty} \frac{1 - (\frac{1}{x^{2}})}{x^{2} + (\frac{1}{x^{2}}) + k}}} d x$