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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.

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

2 Answers 2

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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 $I_1=\frac{I_1+I_2}2...$ $$I_1=\frac12\int_{\frac{1}{3}}^{3} \frac{\arctan(x)+\arctan(\tfrac1x)}{x^2-x+1}dx\\ =\frac12\int_{\frac{1}{3}}^{3} \frac{\frac{\pi}2}{x^2-x+1}dx\\ =\frac{\pi}4\int_{\frac{1}{3}}^{3} \frac{4}{(2x-1)^2+3}dx\\ =\frac{\pi}{2\sqrt3}\arctan(\tfrac{2x-1}{\sqrt3})\vert_{1/3}^3\\ =\frac\pi{2\sqrt3}(\arctan(\tfrac5{\sqrt3})-\arctan(-\tfrac1{3\sqrt3}))\\ =\frac\pi{2\sqrt3}\arctan(4\sqrt3) $$

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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.

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