IIT JEE Problem too.

since it is of $\frac{0}{0}$ form apply L-hospital and for integral part differentiation apply newton-leibnitz theorem

we get

$\displaystyle \lim_{x\to 0} \frac{x ln(1+x)}{3x^{2}(x^{4}+4)}$

cancelling $x$

$\displaystyle \lim_{x\to 0} \frac{ln(1+x)}{3x(x^4+4)}$

$\displaystyle \lim_{x\to 0} \frac{x-\frac{x^2}{2}+\frac{x^3}{3}-........}{3x(x^4+4)}$

taking $x$ common from numerator

$\displaystyle \lim_{x\to 0} \frac{1-\frac{x}{2}+\frac{x^2}{3}-........}{3(x^4+4)}$

apply the limit $x\to 0$

u get $\frac{1}{12}$