Answer
$$\int^{e^{\pi/4}}_{1}\frac{4dt}{t(1+\ln^2t)}=4\tan^{-1}\frac{\pi}{4}$$
Work Step by Step
$$A=\int^{e^{\pi/4}}_{1}\frac{4dt}{t(1+\ln^2t)}$$
We set $u=\ln t$, which means $$du=\frac{dt}{t}$$
For $t=1$, we have $$u=\ln1=0$$
For $t=e^{\pi/4}$, we have $$u=\ln(e^{\pi/4})=\frac{\pi}{4}$$
Therefore, $$A=4\int_0^{\pi/4}\frac{1}{1+u^2}du=4\tan^{-1}u\Big]_0^{\pi/4}$$ $$A=4(\tan^{-1}\frac{\pi}{4}-\tan^{-1}0)$$ $$A=4\tan^{-1}\frac{\pi}{4}$$
*Note: Do not confuse $\tan\frac{\pi}{4}=1$ and $\tan^{-1}\frac{\pi}{4}\approx38.146^{o}$
