Proving that $\displaystyle \int_0^x \frac{\sin t}{t+1}dt > 0$ for all $x >0$ 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?
 A: You're on the right track.. you also have to show that ${\displaystyle \int_{2k\pi}^x {\sin(t) \over t + 1}\,dt > 0}$ for all $2k\pi < x < 2(k+1)\pi$, since you have 
$$ \int_{0}^x {\sin(t) \over t + 1}\,dt = \sum_{i = 0}^{k-1} \bigg(\int_{2i\pi}^{(2i + 1)\pi} {\sin(t) \over t + 1}\,dt + \int_{(2i+1)\pi}^{(2i + 2)\pi} {\sin(t) \over t + 1}\,dt\bigg) + \int_{2k\pi}^x {\sin(t) \over t + 1}\,dt  $$
You've shown the first sum is positive, but you still have to show the last term is positive too. For that part, I suggest showing that ${\displaystyle \int_{2k\pi}^x {\sin(t) \over t + 1}\,dt}$ increases as $x$ goes from $2k\pi$ to $(2k + 1)\pi$, and then decreases as $x$ goes from $(2k + 1)\pi$ to $(2k + 2)\pi$. From what you've done already, you know it will not decrease all the way to zero.

And now for the "slick trick" solution: Note that the derivative of $1 - \cos(x)$ is $\sin(x)$, and the derivative of ${\displaystyle {1 \over t + 1}}$ is ${\displaystyle -{1 \over (t + 1)^2}}$. So integrating by parts you have
$$\int_{0}^x {\sin(t) \over t + 1}\,dt  = {1 - \cos(t) \over t + 1}\bigg|_{t = 0}^{t = x} + 
\int_0^x {1 - \cos(t) \over (t + 1)^2}\,dt$$
$$= {1 - \cos(x) \over x + 1} + \int_0^x {1 - \cos(t) \over (t + 1)^2}\,dt$$
Since $1 - \cos(t) \geq 0$ for all $t$, the first term is nonnegative. Similarly, the integrand of the second term is nonnegative and thus the resulting integral is positive for $x > 0$.
A: The idea and the execution are fine. I would prefer a small variant.  We want to prove that the integral from $2k\pi$ to $(2k+2)\pi$ is positive.  Look at the integral from $(2k+1)\pi$ to $(2k+2)\pi$. The first half of the interval is fine. For the second half, make the change of variable $t=s+\pi$. Note that $\sin t=-\sin s$. It follows that 
$$\int_{t=(2k+1)\pi}^{(2k+2)\pi} \frac{\sin t}{t+1}dt=-\int_{s=2\pi}^{(2k+1)\pi}\frac{\sin s}{s+\pi+1}ds.$$
Now that the dummy variable $s$ has done its duty, replace it by $t$. Then our integral from $2k\pi$ to $(2k+2)\pi$ is equal to 
$$\int_{t=2k\pi}^{(2k+1)\pi} \left(\frac{\sin t}{t+1}-\frac{\sin t}{t+\pi +1}\right)dt.$$
If we wish, we can rewrite this as
$$\int_{t=2k\pi}^{(2k+1)\pi} \frac{\pi \sin t}{(t+1)(t+\pi +1)}dt.$$
The integrand  is non-negative, and the result follows.
One reason I prefer this version of your argument is that now it is easy to get a reasonable estimate of the integral from $2k\pi$ to $(2k+2)\pi$. 
