Prove the inequality: $\int_0^2 \frac{1}{2+\arctan x} dx \geq \ln 2$ I have started taking calculus 2 class without any specific knowledge of calculus 1. Our current topic is concerning definite integral. Could you please help to explain how I can evaluate the following inequality, also suggest which topics do I have to learn in order to catch up with indefinite integral.
$$\int_0^2 \frac{1}{2+\arctan x} dx \geq \ln 2$$
 A: There are two important things to remember to start with here:


*

*$\arctan(0)=0$

*The derivative of $\arctan(x)$ with respect to $x$ is $\frac{1}{x^2+1}$ (Something you will have learned from differential calculus) which we can notice that for all values of $x$ is a positive number strictly less than $1$ with the exception of when $x=0$ where it is identically equal to $1$.
These two facts together along with noting the derivative of $f(x)=x$ is identically equal to $1$ show that on the interval $(0,2)$ you have that $x$ is always strictly larger than $\arctan(x)$
Using this, we find that on the interval $(0,2)$ we have that $\frac{1}{2+\arctan(x)}$ is always strictly larger than $\frac{1}{2+x}$ (since we are dividing by a smaller amount) from which it follows that $\int_0^2 \frac{1}{2+\arctan(x)}dx \geq \int_0^2\frac{1}{2+x}dx$
Finally, correctly evaluating the integral on the right yields the value of $\ln(2)$ which when replaced in the inequality completes the proof.
A: As $\arctan x<x$, 
$$\frac1{2+\arctan x}>\frac1{2+x}$$
and
$$\int_0^2\frac1{2+\arctan x}>\int_0^2\frac1{2+x}=\log 2.$$

If need be, the first inequality can be established as follows:
$$\frac1{1+t^2}<1$$ and
$$\int_0^x\frac1{1+t^2}\,dt<\int_0^x dt.$$
