Answer
0
Work Step by Step
Let
\[I=\int_{-1}^{1}\frac{\sin x}{1+x^2}\:dx\]
Here \[f(x)=\frac{\sin x}{1+x^2}\]
\[\Rightarrow f(-x)=\frac{\sin (-x)}{1+(-x)^2}\]
\[\Rightarrow f(-x)=-\frac{\sin x}{1+x^2}\]
\[\Rightarrow f(-x)=-f(x)\]
$\Rightarrow f(x)$ is odd function
We will use the property ,if $f(x)$ is odd function
\[\int_{-a}^{a}f(x)dx=0 \;\;\;...(1)\]
Using (1) \[I=0\]
Hence $I=0$
