How to evaluate the following integral: $ \int_{0}^{\infty} \frac{dx}{(x + 1)\sqrt{(x + 1)^{3}a^2 + (1 - a^2)}}$?

Question

I have the integral $$ \int_{0}^{\infty} \frac{dx}{(x + 1)\sqrt{(x + 1)^{3}a^2 + (1 - a^2)}}$$ How to evaluate it? I tried to reduce it to beta-function, but I failed.

Answer 1

You an rewrite the integral as $$I = \int_1^{\infty} \frac{dx}{x\sqrt{x^{3}a^2 + (1 - a^2)}}$$ Now let $t = \sqrt{x^{3}a^2 + (1 - a^2)}$, i.e., $t^2 = x^3a^2 + (1-a^2)$. Hence, we have $$2tdt = 3a^2x^2dx \implies x^2 dx = \dfrac{2t}{3a^2}dt$$ Hence, $$I = \int_1^{\infty} \dfrac{2tdt}{3a^2x^3t} = \int_1^{\infty} \dfrac{2tdt}{3(t^2+a^2-1)t} = \dfrac23 \int_1^{\infty} \dfrac{dt}{t^2+a^2-1}$$ I trust you can finish it from here.