Let D (a) be the given integral then by the rule of differentiation under the integral sign

$\dfrac {dI}{da}= \int\limits_0^{\infty}\dfrac {\partial b}{\partial a}dx=\int\limits_0^{\infty}\dfrac 1{x^2}.\dfrac 1{11+ax^2}x^2dx\\ =\int\limits_0^{\infty}\dfrac 1{11+ax^2}dx\\ =\dfrac 1a\int\limits_0^{\infty}\dfrac 1{\frac {11}a+x^2}\\ =\dfrac 1a.\sqrt{\dfrac a{11}}\Bigg[\tan^{-1}\dfrac {x}{\sqrt{\frac {11}a}}\Bigg]_0^{\infty}\\ =\Bigg[\dfrac 1{\sqrt{a\times 11}}\times \dfrac \pi2\\ \dfrac {dI}{da}=\dfrac \pi2\times \dfrac 1{11a}$

Integrating with respect to a

We get

$$I(0)= \int\limits\dfrac {\pi}{2\sqrt{11}}\times \dfrac 1{\sqrt a}da$$ $ =\dfrac {\pi}{2\sqrt{11}}\times 1\sqrt a da\\ =\dfrac {\pi}{2\sqrt{11}}\times \dfrac {\sqrt a}{\frac 12}+c \\ = \dfrac {\pi}{\sqrt{11}}\sqrt a+c ----\gt (1)\\ I(0)=\int\limits_0^{\infty}\dfrac {\log(11+0)}{x^2}dx \\ I(0)=\dfrac {\log (11)}{-3x^3}$

Put $a = 0$ in eq (1) we get

$I(0) = 0 + c\\ \therefore -\dfrac {\log 11}{3x^3}=c\\ \therefore I(a)=\dfrac {\pi}{\sqrt {11}}\sqrt a-\dfrac 1{3x^3}\log 11$