Evaluate:
$$ \int_0^1 \frac{\arctan \dfrac{x}{x+1}}{\arctan \dfrac{1+2x-2x^2}{2}}\,\,{\rm d}x$$
Solution:
Let $\displaystyle f(x)=\frac{\arctan \dfrac{x}{x+1}}{\arctan \dfrac{1+2x-2x^2}{2} }=\frac{\arctan \dfrac{x}{x+1}}{\arctan \dfrac{1+2x(1-x)}{2}} $.
We note that:
$$ f(x)+f(1-x)=\frac{\arctan \dfrac{x}{x+1}}{\arctan \dfrac{1+2x(1-x)}{2}} + \frac{\arctan \dfrac{1-x}{2-x}}{\arctan \dfrac{1+2(1-x)x}{2}}= \cdots =1 $$
by making use of the formula: $\displaystyle \tan \left ( x+y \right )= \frac{\tan x +\tan y}{1-\tan x \tan y}$.
Therefore, if $I$ denotes the given integral we have that:
$$2I=\int_{0}^{1}f(x)\,{\rm d}x + \int_{0}^{1}f(1-x)\, {\rm d}x = \int_{0}^{1}\, {\rm d}x \Rightarrow I = \frac{1}{2}$$
and the exercise comes to an end.
$$ \int_0^1 \frac{\arctan \dfrac{x}{x+1}}{\arctan \dfrac{1+2x-2x^2}{2}}\,\,{\rm d}x$$
Solution:
Let $\displaystyle f(x)=\frac{\arctan \dfrac{x}{x+1}}{\arctan \dfrac{1+2x-2x^2}{2} }=\frac{\arctan \dfrac{x}{x+1}}{\arctan \dfrac{1+2x(1-x)}{2}} $.
We note that:
$$ f(x)+f(1-x)=\frac{\arctan \dfrac{x}{x+1}}{\arctan \dfrac{1+2x(1-x)}{2}} + \frac{\arctan \dfrac{1-x}{2-x}}{\arctan \dfrac{1+2(1-x)x}{2}}= \cdots =1 $$
by making use of the formula: $\displaystyle \tan \left ( x+y \right )= \frac{\tan x +\tan y}{1-\tan x \tan y}$.
Therefore, if $I$ denotes the given integral we have that:
$$2I=\int_{0}^{1}f(x)\,{\rm d}x + \int_{0}^{1}f(1-x)\, {\rm d}x = \int_{0}^{1}\, {\rm d}x \Rightarrow I = \frac{1}{2}$$
and the exercise comes to an end.
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