<s>[INST] How to compute the integral $(1+t^2) / (1-t^2) $ compute $\displaystyle\int \frac{1+t^2}{1-t^2}dt$
I tried splittting them to two parts, and computed $\displaystyle\int \frac{1}{1-t^2}dt$ using trig sub, but i don't know how to compute the second part.
 [/INST] Use the trig substitution as you did before, but don't stop there! use $$t=\tan{\frac{\theta}{2}}$$
This will give you the following identities:
$$1+t^2=\sec^2{\frac{\theta}{2}}$$
Thus
$$\frac{1}{1+t^2}=\cos^2{\frac{\theta}{2}}$$
and 
$$1-\frac{1}{1+t^2}=1-\cos^2{\frac{\theta}{2}}=\sin^2{\frac{\theta}{2}}$$
Giving
$$\frac{1+t^2-1}{1+t^2}=\frac{t^2}{1+t^2}=\sin^2{\frac{\theta}{2}}$$
So overall (use line 3 and 5!)
$$\frac{1-t^2}{1+t^2}=\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}$$
Using Double Angle Formula this gives 
$$\frac{1-t^2}{1+t^2}=\cos{\theta}$$
and
$$dt = \frac{1}{2}\sec^2{\frac{\theta}{2}}$$
You may not need it all but have a go now :D
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