Evaluating the integral $\int_{1/3}^3 \frac{\tan^{-1}(x) } {1+x^2-x} \, dx$

Question

Would like some help in evaluating the integral $$\int_{1/3}^3 \frac{\tan^{-1}x}{ 1-x+x^2} \, dx$$

Answer 1

Changing $u=\frac{1}{x}$, we get:

$$\int_{1/3}^3 \frac{\tan^{-1}x}{ 1-x+x^2} \, dx=\int_{1/3}^3 \frac{\tan^{-1}\frac{1}{u}}{ 1-u+u^2} \, du$$

Therefore, using $\tan^{-1}x+\tan^{-1} \frac{1}{x}=\frac{\pi}{2}$ for $x>0$, we get:

$$ \begin{aligned} \int_{1/3}^3 \frac{\tan^{-1}x}{ 1-x+x^2} \, dx &= \frac{1}{2}\int_{1/3}^3 \frac{\tan^{-1}x+\tan^{-1}x}{ 1-x+x^2} \, dx\\ &= \frac{\pi}{4}\int_{1/3}^3 \frac{1}{ 1-x+x^2} \, dx \end{aligned} $$

Can you end it now?

Answer 2

$$ \int_{1/3}^3 \frac{\tan^{-1}(x)}{ 1-x+x^2} \, dx = \int_{1/3}^3 \frac{\tan^{-1}(x)}{ (1/x)-1+x} \cdot \frac{dx} x $$ \begin{align} & x = 1/u \\[6pt] & dx = -du/u^2 \\[6pt] & \frac{dx} x = -\frac {du} u \\[6pt] & \text{As } x \text{ goes from } 1/3 \text{ to } 3, \\ & u \text{ goes from } 3 \text{ to } 1/3. \\[6pt] & \tan^{-1} x = \frac\pi 2 - \tan^{-1} u \end{align} So the integral becomes \begin{align} & \int_3^{1/3} \frac{\frac\pi2 - \tan^{-1}(u)}{u - 1 + (1/u)} \cdot\frac{-du} u \\[8pt] = {} & \frac \pi 2\int_{1/3}^3 \frac{-du}{u^2-u+1} + \int_{1/3}^3 \frac{\tan^{-1}(u)}{u^2-u+1} \, du \end{align} So $$ \Big(\text{original integral} \Big) = -\frac \pi 2 \int_{1/3} \frac{du}{u^2-u+1} +\Big(\text{original integral} \Big) $$ Therefore \begin{align} 2\times\Big(\text{original integral}\Big) & = \frac{-\pi}2\int_{1/3}^3 \frac{du}{u^2 - u + 1} \\[8pt] & = \frac{-\pi} 2\int_{1/3}^3 \frac{du}{\left( u - \frac 1 2 \right)^2 + \frac 3 4} \end{align} etc.

Answer 3

Hint. One may recall that $$ \tan^{-1}\frac1x=\frac \pi2-\tan^{-1}x,\quad x>0, $$ then, performing the change of variable $\displaystyle x \to \frac 1x$, gives easily $$ I:=\int_{1/3}^3 \frac{\tan^{-1}x}{ 1-x+x^2} \, dx=\int_{1/3}^3 \frac{\frac \pi2-\tan^{-1}x}{ 1-\frac1x+\frac1{x^2}} \, \frac{dx}{x^2}=\frac \pi2\int_{1/3}^3 \frac{1}{ 1-x+x^2} \, dx-I. $$