Answer
$$\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{1+4cotx}{4-cotx}dx=ln(\frac{4\sqrt{2}}{3})$$
Work Step by Step
$$\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{1+4cotx}{4-cotx}dx=\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{1+4cotx}{4-cotx}\frac{sinx}{sinx}dx$$
$$=\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{sinx+4cosx}{4sinx-cosx}dx$$
$$let\ t=4sinx-cosx,dt=(4cosx+sinx)dx$$
$$\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{sinx+4cosx}{4sinx-cosx}dx\int_{\frac{3\sqrt{2}}{2}}^{4}\frac{1}{t}dt$$
$$=\left [ ln\,t \right ]_{\frac{3\sqrt{2}}{2}}^{4}=ln(\frac{4\sqrt{2}}{3})$$
