Finding the limit of a definite integral

Question

Evaluate: $$ \operatorname*{Lim}_{x \to + \infty} \frac{\mathrm{d}}{\mathrm{d}x} \int_{2 \sin \frac{1}{x}}^{3 \sqrt{x}} \frac{3 t^4 + 1}{(t-3)(t^2 + 3)} \,\mathrm{d}t $$

I have tried applying the Newton-Leb rule to the integral which becomes a really big equation and the there $\mathrm{d}/\mathrm{d}x$ of that. I was thinking of applying L’Hospital rule but I’m not sure if I’ll get a right answer with such a big equation. Is there any other way to solve it?

Answer 1

As you know that according to Leibnitz theorem for integrls $$\frac{d}{dx}\,\int_{a(x)}^{b(x)} f(t)\,dt=f(b(x))\, b'(x)-f(a(x)) \,a'(x)$$

$$\frac{dI}{dx}= \lim_{x \to \infty}\left(\frac{3^5 x^2+1}{9(\sqrt{x}-1)(3x^2+1)}\cdot\frac{3}{2\sqrt{x}}-\frac{3\cdot2^4\sin^4(1/x)+1}{(2\sin(1/x)-3)(4\sin^2(1/x)+3)}\cdot2 \cos(1/x)(-\frac{1}{x^2})\right)$$