Answer to Question #79975 in Calculus for Anand

Question

Using MVT theorem to prove that

x/(1+x^2) < tan (^-1) x < x

Answer 1

Prove the inequality for $x>0$ (otherwise it doesn’t hold)

$\int\limits_{0}^{x}{\frac{1}{1+{{t}^{2}}}dt}={{\tan }^{-1}}x$

By MVT

$\int\limits_{0}^{x}{\frac{1}{1+{{t}^{2}}}dt}=\frac{1}{1+{{\xi }^{2}}}x,\xi \in \left( 0,x \right)$

We have

$\frac{1}{1+{{\xi }^{2}}}\cdot x<1\cdot x=x \\ \frac{1}{1+{{\xi }^{2}}}\cdot x>\frac{x}{1+{{x}^{2}}} \\$

Thus

$\frac{x}{1+{{x}^{2}}}<{{\tan }^{-1}}x<x$