Integrating a Tricky Rational Function with Substitution

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Homework Statement


\displaystyle \int^∞_0 \dfrac{dx}{a^2 + \left(x-\frac{1}{x} \right)^2} a>=2


The Attempt at a Solution



\displaystyle \int^∞_0 \dfrac{x^2 dx}{x^2a^2 + (x^2-1)^2}
 
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utkarshakash said:

Homework Statement


\displaystyle \int^∞_0 \dfrac{dx}{a^2 + \left(x-\frac{1}{x} \right)^2} a>=2


The Attempt at a Solution



\displaystyle \int^∞_0 \dfrac{x^2 dx}{x^2a^2 + (x^2-1)^2}

Put ##\displaystyle x=\frac{1}{t}## in the given integral. :wink:
 
utkarshakash said:

Homework Statement


\displaystyle \int^∞_0 \dfrac{dx}{a^2 + \left(x-\frac{1}{x} \right)^2} a>=2


The Attempt at a Solution



\displaystyle \int^∞_0 \dfrac{x^2 dx}{x^2a^2 + (x^2-1)^2}

Not sure about Pranav's hint but here is how I would have done it:

Making a tricky substitution of,

1/t = x-1/x

OR

In your attempt at solution, expand the denominator, then write numerator as x2-1+1, break the denominator, then in each integrand, divide both sides by x2, try making the denominator the perfect square, then in term like Y2 in the denominator, let Y=t...etc..
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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