How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$

Consider the substitution $x=4\sin y$. Then $dx=4\cos y\ dy$, and
$$\frac 1{(16-x^2)^2} = \frac 1{(16\cos^2 y)^2} = \frac 1{(4\cos y)^4} $$
So,
$$\int\frac 1{(16-x^2)^2}dx = \int\frac{4\cos y}{(4\cos y)^4} dy$$
So, finally you need $\displaystyle\int\frac1{\cos^3}$.
