proove by integraion that (1/2)arctan(2x/1-2X) and arctan(1/x) are both integrals of the same function

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proove by integraion that (1/2)arctan(2x/1-2X) and arctan(1/x) are both integrals of the same function

Mathematics
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think u mean by differentiaion
ok help me out please
no it cant be by differentiation as we r tryin to roll back the two integrals but ignore the c in F(x) + \[c \int\limits_{?}^{?}\]

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F(x) + c
so the first is \[\frac{1}{2}\tan^{-1} (\frac{2x}{1-2x})\]
dont think its possbile they must only differ by a constant
[S] [tan^-1 (2x/1-2X)]/2 = [S] tan^-1(1/x) right?
prob try some sum of angle formulas, possibly, but I dnt think it will make a difference
thanx guys.....yes they have to differ by a constant.....its time consuming whew......how can i master induction of inequalities?
huh, maybe diff of angles formula
tan^-1(1/x) = u -1/x^2 --------- dx = du perhaps 1 + 1/x^2
tan(u) = 1/x is what i was thinking lol; lost track mid type
the difference is an angle of \[\pi/2\]
Are you sure you've stated the problem correctly? If you meant \[1/2\arctan(2x/1-x^2)\] ie x^2 rather than 2x on the bottom line, then the two expressions are identical, because tan2x = 2tan(x/2)/(1+tan^2(x/2)
eash man my bad ey guess you are a geenie
what, \[\frac{1}{2}\tan^{-1} (\frac{2x}{1-x^2}) \neq \tan^{-1} (\frac{1}{x})\]
You are quite right. Again! I'm not there yet, but using brute force with Mathematica might give us some ideas. I checked 2x on the bottom line as well as x^2: Integrate[ArcTan[1/x], x] == x*ArcCot[x] + Log[1 + x^2]/2 Integrate 1/2tan^-1(2x/(1-x^2) == (x*ArcTan[(2*x)/(1 - x^2)] - Log[1 + x^2])/2 Integrate 1/2tan^-1(2x/(1-2x) == (ArcTan[1 - 4*x]/4 + x*ArcTan[(2*x)/(1 - 2*x)] - Log[1 - 4*x + 8*x^2]/8)/2
^ I wouldnt worry about thinking about it, it was a bad question to begin with, I am very sure there was a problem with it

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