Question:
Write a value of $\int \frac{\sec ^{2} x}{(5+\tan x)^{4}} d x$
Solution:
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 $t=\tan x$
$y=-\frac{1}{3(5+\tan x)^{3}}+c$
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