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Need some help with this integral

$$I (\alpha) = \int_1^\infty {\arctan(\alpha x) \over x^2\sqrt{x^2-1}} dx$$

Taking the first derivative with respect to $\alpha$

$$I' (\alpha) = \int_1^\infty { dx\over (1+\alpha^2 x^2) x\sqrt{x^2-1} }$$

What transformations to use in order to solve $I'(\alpha)$?

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Substitute $$u=\sqrt{x^2-1}\implies du=\frac{x}{\sqrt{x^2-1}}dx\implies dx=\frac{\sqrt{x^2-1}}{x}du$$ Then $$\int { dx\over (1+\alpha^2 x^2) x\sqrt{x^2-1} }=\int { du\over x^2(1+\alpha^2 x^2)}=\int { du\over (u^2+1)(a^2u^2+a^2+1) }$$ Perform partial fraction decomposition $$\int { du\over (u^2+1)(a^2u^2+a^2+1) }=\int\frac{du}{u^2+1}-a^2\int\frac{du}{a^2u^2+a^2+1}$$ Sure you know that $$\int\frac{du}{u^2+1}=\arctan(u)+C$$ To solve for $$\int\frac{du}{a^2u^2+a^2+1}$$ Use substitution $$v=\frac{au}{\sqrt{a^2+1}}\implies du=\frac{a^2+1}{a}$$ $$\int\frac{du}{a^2u^2+a^2+1}=\int\frac{\sqrt{a^2+1}dv}{a((a^2+1)v^2+a^2+1)}=\frac{1}{a\sqrt{a^2+1}}\int\frac{dv}{v^2+1}=\frac{\arctan(v)}{a\sqrt{a^2+1}}+C$$ Now plug in back $x$, you would get

$$\int_{1}^{\infty} { dx\over (1+\alpha^2 x^2) x\sqrt{x^2-1} }=\left[\arctan\left(\sqrt{x^2-1}\right)-\frac{a\arctan\left(\frac{a\sqrt{x^2-1}}{a^2+1}\right)}{\sqrt{a^2+1}}\right]_{1}^{\infty}$$ I think you can handle the rest of the calculation.

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  • $\begingroup$ Perfect, super clear and so easy now! Thank you so much for your help!!! $\endgroup$ – Kat Jan 21 at 11:10
  • $\begingroup$ You are welcome $\endgroup$ – Larry Jan 21 at 13:46

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