I am asked to solve this question:
'**Which one is the higher order infinitesimals when** $\mathbf{x \rightarrow 0 \, \int\_0^{\sin x} (1+t)^{\frac{1}{t}} dt}$' **or** $\mathbf{x^2}$.

I know that:I have to solve $\lim\_{x \rightarrow 0}\frac{\int\_0^{\sin x} (1+t)^{\frac{1}{t}} dt}{x}$ by using L 'Hospital's rule and compare the result with $x^2$
But I cant find the derivative.

I have recited that the derivative of $\int\_{φ(x)}^{ψ(x)}f(t)dt$ (if x is not in f(t)) is $f[φ(x)]φ'(x)-f[(ψ)]ψ'(x)$, without actually knowing how it comes.

But for now, this equation seems impossible since I cannot get $(1+0)^{\frac{1}{0}}$, I don't think $\lim (1+0)^{\frac{1}{0}}$ should be right, since the hint said the $\int\_0^{\sin x} (1+t)^{\frac{1}{t}} dt$ ~ $ex$

So what should I do? Is the equation I have recited succeed all the time?