$$\textbf{Part 2}$$

$$\int_{c}\underline{F}.d\underline{r} = \int_{c_1}\underline{F}.d\underline{r} + \int_{c_2}\underline{F}.d\underline{r} + \int_{c_3}\underline{F}.d\underline{r} + \int_{c_4}\underline{F}.d\underline{r}$$

c1:

$$\underline{r}(t) = \langle t , 1 \rangle , t \in [1,-1]$$

$$\underline{r}'(t) = \langle 1 , 0 \rangle$$

$$\int_{c_1}\underline{F}.d\underline{r} = \int_{1}^{-1} \langle \frac{-1}{t^2+1} , \frac{t}{t^2+1} \rangle. \langle 1 , 0 \rangle dt$$

$$= \int_{1}^{-1} \frac{-1}{t^2+1} dt = -\tan^{-1}{t}\Biggr|_{1}^{-1}$$

$$= \frac{\pi}{2}$$

c2:

$$\underline{r}(t) = \langle -1 , t \rangle , t \in [1,-1]$$

$$\underline{r}'(t) = \langle 0 , 1 \rangle$$

$$\int_{c_2}\underline{F}.d\underline{r} = \int_{1}^{-1} \langle \frac{-t}{t^2+1} , \frac{-1}{t^2+1} \rangle. \langle 0 , 1 \rangle dt$$

$$= \int_{1}^{-1} \frac{-1}{t^2+1} dt = -\tan^{-1}{t}\Biggr|_{1}^{-1}$$

$$= \frac{\pi}{2}$$

c3:

$$\underline{r}(t) = \langle t , -1 \rangle , t \in [-1,1]$$

$$\underline{r}'(t) = \langle 1 , 0 \rangle$$

$$\int_{c_3}\underline{F}.d\underline{r} = \int_{-1}^{1} \langle \frac{1}{t^2+1} , \frac{t}{t^2+1} \rangle. \langle 1 , 0 \rangle dt$$

$$= \int_{-1}^{1} \frac{1}{t^2+1} dt = \tan^{-1}{t}\Biggr|_{-1}^{1}$$

$$= \frac{\pi}{2}$$

c4:

$$\underline{r}(t) = \langle 1 , t \rangle , t \in [-1,1]$$

$$\underline{r}'(t) = \langle 0 , 1 \rangle$$

$$\int_{c_4}\underline{F}.d\underline{r} = \int_{-1}^{1} \langle \frac{-t}{t^2+1} , \frac{1}{t^2+1} \rangle. \langle 0 , 1 \rangle dt$$

$$= \int_{-1}^{1} \frac{1}{t^2+1} dt = \tan^{-1}{t}\Biggr|_{-1}^{1}$$

$$= \frac{\pi}{2}$$

$$\int_{c}\underline{F}.d\underline{r} = 4 \times \frac{\pi}{2} = 2\pi$$

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