$$\int^{3}_{-1}\bigg(\tan^{-1}\bigg(\frac{x}{x^2+1}\bigg)+\tan^{-1}\bigg(\frac{x^2+1}{x}\bigg)\bigg)dx$$
what i try
from $$\tan^{-1}(t)+\tan^{-1}\bigg(\frac{1}{t}\bigg)=\frac{\pi}{2}.$$
$$\int^{3}_{-1}\tan^{-1}\bigg(\bigg(\frac{x}{x^2+1}\bigg)+\tan^{-1}\bigg(\frac{x^2+1}{x}\bigg)\bigg)dx=\int^{3}_{-1}\frac{\pi}{2}=2\pi$$
But answer is $\pi$
How do i solve it Help me please
HINT: On $[-1,1]$ the function is odd. (I am assuming that the parentheses are like in your solution).
Since $t= \frac{X}{1+X^2}$ is an odd function for X$\in $[-1,1] $$ \int_{-1}^{1} (tan^{-1}(t(X))+tan^{-1}(\frac{1}{t(X)}))dX=0$$ $$\int_{1}^{3} (π/2)dX=2.(π/2)=π$$
Note the identity below is sign-dependent
$$\tan^{-1}(t)+\tan^{-1}\bigg(\frac{1}{t}\bigg)=\text{sgn}(t)\cdot\frac{\pi}{2}$$
Thus $$\int^{3}_{-1}\bigg(\tan^{-1}\bigg(\frac{x}{x^2+1}\bigg)+\tan^{-1}\bigg(\frac{x^2+1}{x}\bigg)\bigg)dx =\int_{-1}^0 \left(-\frac\pi2\right)dx +\int_{0}^3 \left(\frac\pi2\right)dx=\pi$$
(, as is suggested by your attempt. – Przemysław Scherwentke Jan 02 '20 at 06:34