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.

$$\int_0^2 \frac{1}{2+\arctan x} dx \geq \ln 2$$

  • 3
    $\begingroup$ Can you compare $\arctan(x)$ to $x$? On the interval $(0,2)$ which is bigger? Is $\arctan(x)$ always bigger or is $x$ always bigger or do they some times change which is bigger over that interval? Then using this knowledge, which is bigger over that interval? $\frac{1}{2+\arctan(x)}$ or $\frac{1}{2+x}$? Can you evaluate $\int\limits_0^2\frac{1}{2+x}dx$ and use that result to compare to $\int\limits_0^2\frac{1}{2+\arctan(x)}dx$? $\endgroup$
    – JMoravitz
    Feb 5, 2020 at 17:12
  • $\begingroup$ Yes, I think that will work. Thanks! $\endgroup$
    – Lex
    Feb 5, 2020 at 17:19
  • $\begingroup$ @FriendlyFred, please, specify your titles in future. It is too general. $\endgroup$
    – PinkyWay
    Feb 5, 2020 at 17:20
  • $\begingroup$ Okay, this is the first question I am posting, so will be more specific in the future) $\endgroup$
    – Lex
    Feb 5, 2020 at 17:23
  • $\begingroup$ Welcome to Math.SE, I hope you stay around and contribute to the site. $\endgroup$
    – gt6989b
    Feb 5, 2020 at 17:28

2 Answers 2


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.

  • $\begingroup$ Yes, of course! It's the best way here. +1 $\endgroup$ Feb 5, 2020 at 17:27
  • $\begingroup$ Thanks for such detailed explanation! $\endgroup$
    – Lex
    Feb 5, 2020 at 17:35

As $\arctan x<x$,

$$\frac1{2+\arctan x}>\frac1{2+x}$$


$$\int_0^2\frac1{2+\arctan x}>\int_0^2\frac1{2+x}=\log 2.$$

If need be, the first inequality can be established as follows:

$$\frac1{1+t^2}<1$$ and

$$\int_0^x\frac1{1+t^2}\,dt<\int_0^x dt.$$

  • $\begingroup$ Thank you it really helped! $\endgroup$
    – Lex
    Feb 5, 2020 at 17:35

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