f(x) = int {left( {2 - frac{1}{{1 + {x^2}}} - | vedclass

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(C) To determine the monotonicity of $f(x)$,we find its derivative $f'(x)$.Given $f(x) = \int {\left( {2 - \frac{1}{{1 + {x^2}}} - \frac{1}{{\sqrt {1

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increasing in $(-\infty, \infty)$

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(C) To determine the monotonicity of $f(x)$,we find its derivative $f'(x)$.Given $f(x) = \int {\left( {2 - \frac{1}{{1 + {x^2}}} - \frac{1}{{\sqrt {1 + {x^2}} }}} ight)} \,dx$,by the Fundamental Theorem of Calculus,$f'(x) = 2 - \frac{1}{{1 + {x^2}}} - \frac{1}{{\sqrt {1 + {x^2}} }}$.We analyze the sign of $f'(x)$ for all $x \in \mathbb{R}$.$f'(x) = \frac{2(1 + {x^2}) - 1 - \sqrt{1 + {x^2}}}{1 + {x^2}} = \frac{2{x^2} + 1 - \sqrt{1 + {x^2}}}{1 + {x^2}}$.Let $u = \sqrt{1 + {x^2}}$. Since ${x^2} \ge 0$,we have $u \ge 1$. Then ${x^2} = {u^2} - 1$.Substituting this into the numerator: $2({u^2} - 1) + 1 - u = 2{u^2} - u - 1 = (2u + 1)(u - 1)$.Since $u \ge 1$,$(u - 1) \ge 0$ and $(2u + 1) > 0$,so the numerator is $\ge 0$.Specifically,$f'(x) > 0$ for all $x 
eq 0$,and $f'(0) = 0$.Since $f'(x) \ge 0$ for all $x \in \mathbb{R}$ and is not identically zero on any interval,$f(x)$ is strictly increasing on $(-\infty, \infty)$.

$f(x) = \int {\left( {2 - \frac{1}{{1 + {x^2}}} - \frac{1}{{\sqrt {1 + {x^2}} }}} \right)} \,dx$. Then $f$ is:

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