Hint required : Why is the integral $\int_0^x \frac{\sin(t)}{1+t}\mathrm{d}t$ positive? $\sin(t)$ is continuous on $[0,x]$ and $\frac{1}{1+t}$ is continuous on $[0,x]$ so $\frac{\sin(t)}{1+t}$ is continuous on $[0,x]$ so the function is integrable.
How do I proceed? What partition should I consider ? 
Edit : We haven't done any properties of the integral so far except the basic definition, that continuous functions are integrable, and that one may construct a sequence of partitions, and calculate $lim \ U(f,P_n)$ or $lim \ L(f,P_n)$ which must be equal to the integral ( I am not clear whether the mesh tends to zero or not in the last one, since technically, we've done Darboux integrals where there is no concept of mesh.. )
 A: Here is an outline that you can follow. The key idea is to draw the graph and recognize that the area contained within each arc of the function is decreasing, and the first area is positive.



*

*Conclude that for all $x \in [0, \pi]$, $\int_0^x \frac{\sin t}{1 + t} \, dt > 0$. 

*Show that for each integer $k \ge 0$, 
$$\int_{k \pi}^{(k + 1) \pi} \left|\frac{\sin t}{1 + t}\right| \, dt > \int_{(k + 1)\pi}^{(k + 2)\pi} \left|\frac{\sin t}{1 + t}\right| \, dt.$$

*Think about for which values of $k$ you get a positive integrand, and for which values of $k$ you get a negative one.
A: For any $a>0$ and any $x>0$ we have
$$ e^{ax}-\cos(x)-a\sin(x) \geq (1-\cos x)+a(x-\sin x) \geq 0 $$
such that 
$$ \int_{0}^{x}\sin(t)e^{-at}\,dt \geq 0 $$
ensures the non-negativity of the Laplace transform of $\mathbb{1}_{(0,x)}(u)\sin(u)$. The inverse Laplace transform of $\frac{1}{1+u}$ is also non-negative, hence the given integral is non-negative for any $x>0$.
A: I accidentally worked on that problem today. I'm surprised by the coincidence. Here I share my notes.


