We have $f:\mathbb{R}\rightarrow\mathbb{R},\:f(x)=x^2+x\sin(x)$, and we need to find intervals of monotonicity. Here is all my steps:
$f'(x)=2x+x\cos(x)+\sin(x)$
$f'(x)=0 \Rightarrow x=0$ the only solution.
Now I need to find where $f'$ is positive and negative. I don't want to put value for $f'$ to find the sign, I want another method. So I tried to differentiate the function again to see if $f'$ is increasing or decreasing: $f''(x)=2-x\sin(x) + \cos(x)$ , but I don't know if $f''(x)\geq0$ or $f''(x)\leq0$.
How can I find the sign for $f'$ to determine monotony of $f$ ?
You know that the function $f'(x)$ is a continuous function with only one root. This means that the sign of the function is the same on $[0,\infty)$, and this means that no matter what value of $x>0$ you take, you will find the sign of $f'(x)$ on this interval.
In some cases, it's good to take a particular value since that makes it easier. In your case, it's very easy to calculate $f'(2\pi),$ for example.
Also, you can try to see what the limit of $f'(x)$ is when $x$ becomes large. If the limit is positive (or if it is $\infty$), then the sign of $f'$ is also positive on $[0,\infty)$. In your case, the limit is easy to calculate.