# Calculate Riemann integral

Calculate Riemann integral: $$\int_{\frac{1}{3}}^{3} \frac{\arctan(x)}{x^2-x+1}dx$$

In this assignment, the integrand function does not have an antiderivative. I used the substitution of $$x=\frac{1}{t}$$ and the fact that $$\arctan(x) +\arctan \left( \frac{1}{x} \right) = \frac{\pi}{2} (x \neq 0, x \in \mathbb{R})$$ and reduced the integral from the task to the following form:

$$\int_{\frac{1}{3}}^{3} \frac{\arctan(x)}{x^2-x+1}dx = \int_{3}^{\frac{1}{3}} \frac{\arctan(\frac{1}{t})}{\frac{1}{t^2}-\frac{1}{t}+1} \left(-\frac{1}{t^2} \right) dt = \frac{\pi}{2} \cdot \int_{3}^{\frac{1}{3}} \frac{1}{\frac{1}{t^2}-\frac{1}{t}+1} \left(-\frac{1}{t^2} \right) dt - \int_{3}^{\frac{1}{3}} \frac{\arctan(t)}{\frac{1}{t^2}-\frac{1}{t}+1} \left(-\frac{1}{t^2} \right) dt=$$ $$= \frac{\pi}{2} \cdot \int_{3}^{\frac{1}{3}} \frac{1}{\frac{1}{t^2}-\frac{1}{t}+1} \left(-\frac{1}{t^2} \right) dt - \int_{\frac{1}{3}}^{3} \frac{\arctan(\frac{1}{x})}{x^2-x+1}dx \ \left( * \right)$$

I'm stuck on this. I calculated this integral on a calculator in Wolframalpha and it gives out that the value of the integral $$\int_{\frac{1}{3}}^{3} \frac{\arctan(x)}{x^2-x+1}dx = \frac{\pi}{4} \cdot \int_{3}^{\frac{1}{3}} \frac{1}{\frac{1}{t^2}-\frac{1}{t}+1} \left(-\frac{1}{t^2} \right) dt = \frac{\pi}{4} \cdot \int_{\frac{1}{3}}^{3} \frac{1}{x^2-x+1} dx$$

But I don't understand how to get this result from the step $$\left( * \right)$$ where I left off.

I will be glad if you tell me what to do after $$\left( * \right)$$ or suggest another way to calculate the integral.

• The integral on the LHS also appears on the RHS in $(*)$, so that you gather them together on the left before dividing by 2. Feb 22 at 18:38
• Can you please tell me a little more in detail how I can gather them together on the left before dividing by 2? Feb 22 at 18:43
• Doug got ahead of my explanations with this answer ;) Feb 22 at 18:57
• @Abezhiko, sorry bro. Your insight inspired me.
– Doug
Feb 22 at 20:30

Let $$I_1=\int_{\frac{1}{3}}^{3} \frac{\arctan(x)}{x^2-x+1}dx$$ and $$I_2=\int_{\frac{1}{3}}^{3} \frac{\arctan(\tfrac1x)}{x^2-x+1}dx.$$ In the first equality you showed that $$I_1=I_2.$$ Then, I guess you tried to do the trick $$I_1=\frac{I_1+I_2}2...$$ $$I_1=\frac12\int_{\frac{1}{3}}^{3} \frac{\arctan(x)+\arctan(\tfrac1x)}{x^2-x+1}dx\\ =\frac12\int_{\frac{1}{3}}^{3} \frac{\frac{\pi}2}{x^2-x+1}dx\\ =\frac{\pi}4\int_{\frac{1}{3}}^{3} \frac{4}{(2x-1)^2+3}dx\\ =\frac{\pi}{2\sqrt3}\arctan(\tfrac{2x-1}{\sqrt3})\vert_{1/3}^3\\ =\frac\pi{2\sqrt3}(\arctan(\tfrac5{\sqrt3})-\arctan(-\tfrac1{3\sqrt3}))\\ =\frac\pi{2\sqrt3}\arctan(4\sqrt3)$$

Notice that $$\int_{1/3}^{3}\frac{\arctan(1/x)}{x^2-x+1}dx,\quad t = 1/x$$ $$=\int_{3}^{1/3}\frac{\arctan(t)}{(1/t)^2-(1/t)+1}\frac{-1}{t^2}dt$$ $$=-\int_{3}^{1/3}\frac{\arctan(t)}{1-t+t^2}dt$$ $$=\int_{1/3}^{3}\frac{\arctan(t)}{t^2-t+1}dt.$$

So, the (*) integral is the same as the original.