Write a value of

Let, $\tan x=t$

Differentiating both side with respect to $x$

$\frac{d t}{d x}=(\sec x)^{2} \Rightarrow d t=\sec ^{2} x d x$

$y=\int \frac{d t}{(5+t)^{4}}$

Use formula $\int \frac{1}{(a+t)^{n}} d t=\frac{(a+t)^{-n+1}}{-n+1}$

$y=\frac{(5+t)^{-3}}{-3}+c$

Again, put $t=\tan x$

$y=-\frac{1}{3(5+\tan x)^{3}}+c$