# 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.

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

• 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 $\int\limits_0^2\frac{1}{2+\arctan(x)}dx$? Feb 5, 2020 at 17:12
• Yes, I think that will work. Thanks!
– Lex
Feb 5, 2020 at 17:19
• @FriendlyFred, please, specify your titles in future. It is too general. Feb 5, 2020 at 17:20
• Okay, this is the first question I am posting, so will be more specific in the future)
– Lex
Feb 5, 2020 at 17:23
• Welcome to Math.SE, I hope you stay around and contribute to the site. Feb 5, 2020 at 17:28

• $$\arctan(0)=0$$

• 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.

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

As $$\arctan 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

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

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