How do you evaluate the integral $\int{\dfrac{\sin x}{1+{{x}^{2}}}dx}$ from $-\infty $ to $\infty $?

Hint: We will use the concept of the definite integration to solve the above given integration. We will use the property that $\int\limits_{-a}^{a}{f\left( x \right)dx=0}$ when f(x) is an odd function and $\int\limits_{-a}^{a}{f\left( x \right)dx=}2\int\limits_{0}^{a}{f\left( x \right)dx}$ when f(x) is an even function. So, we will first check whether the function $f\left( x \right)=\dfrac{\sin x}{1+{{x}^{2}}}$ is even or odd and then we will use the above property to integrate.

Complete answer:
We will use the properties of function and definite integration to solve the above integration.
Since, we can see that the limit of integration varies from $-\infty $ to $\infty $. And from properties of definite integration we know that $\int\limits_{-a}^{a}{f\left( x \right)dx=0}$ when f(x) is an odd function and $\int\limits_{-a}^{a}{f\left( x \right)dx=}2\int\limits_{0}^{a}{f\left( x \right)dx}$ when f(x) is an even function.
So, we will first check whether the function $f\left( x \right)=\dfrac{\sin x}{1+{{x}^{2}}}$ is even or odd.
Now, when we put -x in place of x we will get:
$f\left( -x \right)=\dfrac{\sin \left( -x \right)}{1+{{\left( -x \right)}^{2}}}$
Since, we know that $\sin \left( -x \right)=-\sin x$. So, we will get:
$\Rightarrow f\left( -x \right)=\dfrac{-\sin \left( x \right)}{1+{{x}^{2}}}$
$\Rightarrow f\left( -x \right)=-\left( \dfrac{\sin x}{1+{{x}^{2}}} \right)$
Since, we know that $f\left( x \right)=\dfrac{\sin x}{1+{{x}^{2}}}$, so we will get:
$\Rightarrow f\left( -x \right)=-f\left( x \right)$
So, we will say that the function $f\left( x \right)$ is an odd function as $f\left( -x \right)=-f\left( x \right)$ is the property of an odd function.
This implies that we will use the property $\int\limits_{-a}^{a}{f\left( x \right)dx=0}$ when f(x) is an odd function.
So, in $\int\limits_{-\infty }^{\infty }{\dfrac{\sin x}{1+{{x}^{2}}}dx}$ , we have $a=\infty $ and $f\left( x \right)=\dfrac{\sin x}{1+{{x}^{2}}}$, when we compared it with \[\int\limits_{-a}^{a}{f\left( x \right)dx}\].
Since, we have proved above that f(x) is an odd function.
So, we will can say that $\int\limits_{-\infty }^{\infty }{\dfrac{\sin x}{1+{{x}^{2}}}dx}=0$.
Hence, the value of the integral $\int{\dfrac{\sin x}{1+{{x}^{2}}}dx}$ from $-\infty $ to $\infty $ is 0.
This is our required solution.

Note: Students are required to memorize all the formulas of integration and also all the properties of definite integration otherwise they will not be able to solve the above question. Also, we can’t directly integrate the above given integration so it is required to use the property of definite integration.