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