$$\left( {({{\tan }^{ - 1}}y) - x} \right)dy = (1 + {y^2})dx$$

$${{dx} \over {dy}} + {x \over {1 + {y^2}}} = {{{{\tan }^{ - 1}}y} \over {1 + {y^2}}}$$

$$I.F. = {e^{\int {{1 \over {1 + {y^2}}}dy} }} = {e^{{{\tan }^{ - 1}}y}}$$

$$\therefore$$ Solution

$$x.\,{e^{{{\tan }^{ - 1}}y}} = \int {{{{e^{{{\tan }^{ - 1}}y}}{{\tan }^{ - 1}}y} \over {1 + {y^2}}}dy} $$

Let $${e^{{{\tan }^{ - 1}}y}} = t$$

$${{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}} = dt$$

$$ = x{e^{{{\tan }^{ - 1}}y}} = \int {\ln tdt = t\ln t - t + c} $$

$$\therefore$$ $$ = x{e^{{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}}{\tan ^{ - 1}}y - {e^{{{\tan }^{ - 1}}y}} + c$$ ..... (i)

$$\because$$ It passes through (1, 0) $$\Rightarrow$$ c = 2

Now put y = tan1, then

$$ex = e - e + 2$$

$$ \Rightarrow x = {2 \over e}$$