Question $2$

Evaluate the integral.

${\displaystyle \underset{0}{\overset{3}{}}}\frac{4e^{-t}}{1+16e^{-2t}}dt$

$ta{n}^{-1}(\frac{4}{e^3})*ta{n}^{-1}4$

$\frac{1}{4}ta{n}^{-1}4-\frac{1}{4}ta{n}^{-1}(\frac{4}{e^3})$

$ta{n}^{-1}3-ta{n}^{-1}4$

$ta{n}^{-1}4*ta{n}^{-1}(\frac{4}{e^3})$