Answer
a) $\tan^{-1} x=\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \gt 1$
b) $\tan^{-1} x=-\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \lt -1$
Work Step by Step
a) When $ x \gt 1$
$\int_x^{\infty}\dfrac{1}{1+t^2}dt= \int_x^{\infty} (\dfrac{1} {t^2}-\dfrac{1} {t^4}+\dfrac{1} {t^6}+.....) dt $
or, $[\tan^{-1} t]_x^{\infty}=[-\dfrac{1} {t}+\dfrac{1} {3t^3}-\dfrac{1} {5t^5}+.....]_x^{\infty}$
So, $\tan^{-1} x=\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \gt 1$
b) When $ x \lt -1$
$\int_{-\infty}^ {x} \dfrac{1}{1+t^2}dt= \int_{-\infty}^ {x} (\dfrac{1} {t^2}-\dfrac{1} {t^4}+\dfrac{1} {t^6}+.....) dt $
or, $[\tan^{-1} t]_{-\infty}^ {x} =[-\dfrac{1} {t}+\dfrac{1} {3t^3}-\dfrac{1} {5t^5}+.....]\int_{-\infty}^ {x} $
So, $\tan^{-1} x=-\dfrac{\pi}{2}-\dfrac{1} {x}+\dfrac{1} {3x^3}-\dfrac{1} {5x^5}.....; x \lt -1$
