Draw the graph of `y=tan^(-1)((2x)/(1-x^(2)))`

Explanation

We have `y=f(x)=tan^(-1)((2x)/(1-x^(2)))`
Domain of f(x) is R-{-1, 1}.
Let `x=tantheta in (-pi//2,pi//2)`
`rArr" "theta=tan^(-1)x`
`"Now "tan^(-1)((2x)/(1-x^(2)))=tan^(-1)((2tantheta)/(1-tan^(2)theta))`
`=tan^(-1)(tan2theta)`
`=tan^(-1)(tanalpha)" where "alpha in (-pi, pi)`
Now consider the graph of `y=tan^(-1)(tan alpha)," where "alpha in (-pi, pi)`.

Form the graph,
`tan^(-1)((2x)/(1-x^(2)))=tan^(-1)(tan alpha)`
`={{:(alpha+pi,-piltalphalt-pi//2),(alpha, -pi//2ltalphaltpi//2),(alpha-pi,pi//2lealphaltpi):}`
`={{:(2tan^(-1)x+pi,-pilttan^(-1)xlt-pi//2),(2tan^(-1)x, -pi//2lttan^(-1)xltpi//2),(2tan^(-1)x-pi,pi//2lt2tan^(-1)xltpi):}`
`={{:(2tan^(-1)x+pi,-pilttan^(-1)xlt-pi//4),(2tan^(-1)x, -pi//4lttan^(-1)xltpi//4),(2tan^(-1)x-pi,pi//4lt2tan^(-1)xltpi//2):}`
`={{:(2tan^(-1)x+pi, -ooltxlt-1),(2tan^(-1)x,-1ltxlt1),(1tan^(-1)x-pi,1ltxltoo):}`
`tan^(-1)" is an increasing function for "x in R`.
Thus, all branch functions in the above are increasing functions.
`underset(xto-oo)(lim)(tan^(-1)x+pi)=0,underset(xto-1)(lim)(2 tan^(-1)x+pi)=pi/2`
Thus, `tan^(-1)((2x)/(1-x^(2)))` increases form 0 to `pi/2` when x increases from `-oo" to "-1`
`underset(xto-1)(lim)(2tan^(-1)x)=pi/2, underset(xto1)(lim)(2 lim^(-1)x)-pi/2" and "2 tan^(-1)0=0`
Thus, `tan^(-1)((2x)/(1-x^(2)))" increases form "-pi/2" to "pi/2" when x increases from -1 to 1, intersecting the x-axis at x=0"`.
`underset(xto1)(lim)(2tan^(-1)x-[pi)=-pi/2,underset(xtooo)(lim)(2tan^(-1)x-pi)=0`
Thus, `tan^(-1)((2x)/(1-x^(2)))" increases form "-pi/2" to0 when x increases from 1 to "oo.`
From this information, we can draw the graph of `y=tan^(-1)((2x)/(1-x^(2)))` as folows.

Here y = 0 is na anymptote.