Proving that if \(\displaystyle{x}{>}{0}\) then \(\displaystyle{\frac{{{x}-{\left({x}^{{2}}+{1}\right)}{\arctan{{\left({x}\right)}}}}}{{{x}^{{2}}{\left({x}^{{2}}+{1}\right)}}}}\)

Dominique Pace

Dominique Pace

Answered question

2022-03-28

Proving that if x>0 then x(x2+1)arctan(x)x2(x2+1) is less than 0

Answer & Explanation

Jayda Burch

Jayda Burch

Beginner2022-03-29Added 10 answers

We have
x(x2+1)arctanxx2(x2+1)<0x<(x2+1)arctanx
(1+x2)arctanxx>0
This is true iff
ddx[(1+x2)arctanxx]=2xarctanx+1+x21+x21=2xarctanx>0
which is true since both x and arctanx are greater than 0 when x>0. Hence result.

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