Answer
$\mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = \frac{\pi }{2}$
Work Step by Step
Write the vector field ${\bf{F}}\left( {x,y} \right) = \left( {\frac{{ - y}}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right)$ and the path ${\bf{r}} = \left( {x,y} \right)$, such that
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}}$
The segment from $\left( {1,0} \right)$ to $\left( {0,1} \right)$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {1,0} \right) + t\left( {\left( {0,1} \right) - \left( {1,0} \right)} \right)$, ${\ \ \ }$ for $0 \le t \le 1$
${\bf{r}}\left( t \right) = \left( { - t + 1,t} \right)$
So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\frac{{ - t}}{{{{\left( { - t + 1} \right)}^2} + {t^2}}},\frac{{ - t + 1}}{{{{\left( { - t + 1} \right)}^2} + {t^2}}}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( { - 1,1} \right)dt$
Using Eq. (8), the vector line integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = \mathop \smallint \limits_0^1 \left( {\frac{{ - t}}{{{{\left( { - t + 1} \right)}^2} + {t^2}}},\frac{{ - t + 1}}{{{{\left( { - t + 1} \right)}^2} + {t^2}}}} \right)\cdot\left( { - 1,1} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = \mathop \smallint \limits_0^1 \left( {\frac{t}{{{{\left( { - t + 1} \right)}^2} + {t^2}}} + \frac{{ - t + 1}}{{{{\left( { - t + 1} \right)}^2} + {t^2}}}} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \frac{1}{{{{\left( { - t + 1} \right)}^2} + {t^2}}}{\rm{d}}t$
The integral above gives
$\mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = \mathop \smallint \limits_0^1 \frac{1}{{2{t^2} - 2t + 1}}{\rm{d}}t$
Re-write the integral at the right-hand side as
$\mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = \mathop \smallint \limits_0^1 \frac{2}{{4{t^2} - 4t + 2}}{\rm{d}}t$
$\mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = 2\mathop \smallint \limits_0^1 \frac{1}{{{{\left( {2t - 1} \right)}^2} + 1}}{\rm{d}}t$
Let $u = 2t - 1$. So, $du = 2dt$. The integral becomes
$\mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = \mathop \smallint \limits_{ - 1}^1 \frac{1}{{{u^2} + 1}}{\rm{d}}u$
By Theorem 2 of Section 7.8, we know that $\frac{d}{{dx}}{\tan ^{ - 1}} = \frac{1}{{{x^2} + 1}}$. Thus, we obtain
$\mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = {\tan ^{ - 1}}u|_{ - 1}^1 = \frac{\pi }{4} - \left( { - \frac{\pi }{4}} \right) = \frac{\pi }{2}$
So, $\mathop \smallint \limits_C^{} \frac{{ - y{\rm{d}}x + x{\rm{d}}y}}{{{x^2} + {y^2}}} = \frac{\pi }{2}$.
