<s>[INST] Integrate $ \int \frac{2\tan x+1}{\sqrt{\tan^2 x+2\tan x+2}} \, \mathrm dx $ I tried writing $\tan^2 x+2\tan x+2$ as $(\tan x+1)^2+1$ and letting $\tan\theta = \tan x+1$ yields
$$ \int \frac{2\tan\theta-1}{\tan^2\theta-2\tan\theta+2}\,\sec\theta\,\mathrm d\theta $$
Then letting $ t = \tan(\frac{\theta}{2})$ yields
$$ \int\frac{t^2+4t-1}{t^4+2t^3-2t+1}\,\mathrm dt $$
Obviously what's next is factoring out the denominator and doing partial fraction decomposition, but this seems very hard.
I wanted to know if there's an easier method.
 [/INST] Let $t=\tan x$ to rewrite the integral as
$$ I=\int \frac{2\tan x+1}{\sqrt{\tan^2 x+2\tan x+2}} dx
=\int \frac{2t+1}{(1+t^2)\sqrt{t^2+2t+2}} dt\tag1
$$
Note that
$$
\bigg( \tan^{-1} \frac{at-a^{-1}}{\sqrt{t^2+2t+2}}\bigg)’
= \frac{a^{-1} t +ab }{(t^2+b)\sqrt{t^2+2t+2}},\>\>\>\>\>
b= \frac{1+2a^2}{a^2+a^4}$$
Set $b=1$ to obtain $a^4-a^2-1=0$, which leads to $a= \sqrt{\phi}$, with $\phi=\frac{1+\sqrt5}2$ the golden ratio. Hence
$$
g(t) = \frac{\phi^{-1/2} t + \phi^{1/2} }{(t^2+1)\sqrt{t^2+2t+2}} =\bigg( \tan^{-1} \frac{\phi^{1/2}t-\phi^{-1/2}}{\sqrt{t^2+2t+2}}\bigg)’
$$
and similarly
$$
h(t) =\frac{\phi^{-1/2}- \phi^{1/2} t }{(t^2+1)\sqrt{t^2+2t+2}} =\bigg( \tanh^{-1} \frac{\phi^{-1/2}t+\phi^{1/2}}{\sqrt{t^2+2t+2}}\bigg)’
$$
Then, the integrand in (1) can be decomposed as
$$
\frac{2t+1}{(1+t^2)\sqrt{t^2+2t+2}} 
= \phi^{1/2} g(t) - \phi^{-1/2} h(t)
$$
resulting in
$$I= \phi^{1/2} \tan^{-1} \frac{\phi^{1/2}t- \phi^{-1/2}}{\sqrt{t^2+2t+2}}
 - \phi^{-1/2}  \tanh^{-1} \frac{\phi^{-1/2}t+ \phi^{1/2} }{\sqrt{t^2+2t+2}}+C
$$
</s>