Answer
$+\infty$
The given integral is divergent.
Work Step by Step
Simplify the improper integral $\int_0^{\pi/4} \dfrac{\sec^2 x}{1-\tan x} \ dx$
Let us consider that $I=\int_0^{\pi/4} \dfrac{\sec^2 x}{1-\tan x} \ dx=\lim\limits_{z \to \pi/4} \int_0^{z} \dfrac{\sec^2 x}{1-\tan x} \ dx$
Integrate the integral by using limits.
$I=\lim\limits_{z \to \pi/4} [\ln |1-\tan x|]\\=\ln |0|-\ln |1| \\= +\infty$
So, the given integral is divergent.
