I devised this proof that
$$\tag{1} \int_0^x \frac{\sin t}{t+1}dt > 0 \text{ ; } \forall x >0$$
The idea is to prove that the area from $(0,\pi)$ is greater than the absolute value of the negative area in $(\pi, 2\pi)$, and so on, so that the final area is always positive.
$f(x) = \dfrac{\sin x}{x+1}$ is positive if $\sin x >0$ and negative if $\sin x <0$. This is to say
$$f(x) >0 \Leftrightarrow x \in \bigcup_{k=0}^{\infty}(2k\pi,(2k+1)\pi)$$
$$f(x) <0 \Leftrightarrow x \in \bigcup_{k=1}^{\infty}((2k-1)\pi,2k\pi)$$
If we prove that $$\tag{2} |f(x)| > |f(x+\pi)|$$ for every $x$ then we prove $(1)$.
But,
$|f(x)| =\left| \dfrac{\sin x}{x+1} \right|$
$|f(x+\pi)| =\left| \dfrac{\sin (x+\pi)}{x+\pi+1} \right|=\left| \dfrac{\sin x}{x+\pi+1} \right|$
Thus $(2)$ is proven, and we then have that in general,
$$ |f(x+n \pi)| > |f(x+(n+1) \pi)|$$, thus
$$\int\limits_{\left( {2k} \right)\pi }^{\left( {2k + 1} \right)\pi } {\frac{{\sin t}}{{t + 1}}dt} + \int\limits_{\left( {2k + 1} \right)\pi }^{\left( {2k + 2} \right)\pi } {\frac{{\sin t}}{{t + 1}}dt} > 0$$
and then
$$ \int_0^x \frac{\sin t}{t+1}dt > 0 \text{ ; } \forall x >0$$
Is it right? And if it is right - is it understandable?