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

The main question is how to write $16-x^2$ as a square. Note that $\cosh^2 \theta-\sinh^2 \theta = 1$ hence $1-\tanh^2\theta = \text{sech}^2 \theta$ thus $16-16\tanh^2\theta = 16\text{sech}^2 \theta$. This suggests a $x=4\tanh \theta$ substitution. Observe $dx = 4\text{sech}^2(\theta) d\theta$. Putting this all together,
$$ \int \frac{dx}{(16-x^2)^2} = \int \frac{4\text{sech}^2(\theta) d\theta}{(16)^2\text{sech}^4 \theta} = \frac{1}{64}\int \cosh^2 \theta d\theta $$
Now, $\cosh^2 \theta = \frac{1}{2}(\cosh 2 \theta+1)$
$$ \frac{1}{64}\int \cosh^2 \theta d\theta = \frac{1}{128}\int (\cosh 2 \theta +1)d\theta = \frac{1}{256}(\sinh 2 \theta + 2\theta)+c $$
Finally, $\theta = \tanh^{-1}(x/4)$ thus,
$$ \int \frac{dx}{(16-x^2)^2}  = \frac{1}{256}\biggl(\sinh \bigl[2 \tanh^{-1}(x/4)\bigr] + 2\tanh^{-1}(x/4)\biggr)+c $$
Of course you can write the first term as an algebraic function by studying the identities for hyperbolic functions and the given hyperbolic subsitution.
( I like the sine or cosine substitution for this problem, but I include this answer for breadth)
