Evaluting $\int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} \operatorname dx$

$$\int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} dx$$

My try:: $\displaystyle \int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} dx = \int_{1}^{2}\frac{\tan^{-1}x}{\tan^{-1}(x-1)-\tan^{-1}(x-2)}dx$

Now How can i solve after that.

plz help me

Thanks

Solution 1:

After a few simple algebraic manipulations, the original integral turns into $$ \mathcal{I}=\frac{1}{2}\int_{0}^{1}\frac{\frac{\pi}{2}-\arctan\left(\frac{5-x^2}{12}\right)}{\frac{\pi}{2}+\arctan\left(\frac{3+x^2}{12}\right)}\,dx$$ which is pretty simple to approximate numerically, for instance through the Shafer-Fink approximation $\arctan(x)\approx\frac{3x}{1+2\sqrt{1+x^2}}$, leading to $\mathcal{I}\approx 1.108$. On the other hand, a simple closed form seems to be out of reach, since there is no evident symmetry and the substitution $\frac{3+x^2}{12}\mapsto\tan u$ does not simplify the integrand function as one might hope.

Solution 2:

Observe that the quadratic $ x^2 -3x +3 $ has discriminant $$b^2-4ac=(-3)^2-4(1)(3)=9-12=-3$$ Hence no real solutions, which means it never crosses the $x$-axis, and since it opens up, this means it is always positive. Using this, and the following formula for $\arctan \frac{1}{u}$ $$\arctan \frac{1}{u}=\frac{1}{2}\pi - \arctan u, u>0 $$ We can rewrite the denominator of our integral as $$\frac{1}{2}\pi - \arctan (x^2-3x+3)$$ having used $u=x^2-3x+3$ to apply the formula. We now have $$\int_1^2 \frac{\arctan x}{\frac{1}{2} \pi - \arctan (x^2-3x+3)}dx$$ This is as far as I got, and wolfram alpha didn't find a closed-form solution for that.