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.
 A: You can use $\int \frac {1+t^2}{1-t^2}\;dt=\int(-1)\;dt+\int\frac 2{1-t^2}\; dt$ and what you have
A: $$\frac {1+t^2}{1-t^2}=\frac {2-1+t^2}{1-t^2}=\frac {2}{1-t^2}-1$$
Now use partial fractions $$\frac 2{1-t^2}=\frac a{1-t}+\frac b{1+t}$$
A: 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
A: Hint:
$$
\frac{1 + t^2}{1 - t^2} = -1+\frac2{1 - t^2} = -1 - \frac1{t - 1} + \frac1{t + 1}$$
A: $$
\int\frac{1+t^2}{1-t^2}dt=-\int\frac{-1-t^2}{1-t^2}dt=
-\int\frac{-2+1-t^2}{1-t^2}dt=\int\frac{2}{1-t^2}dt-\int 1dt.
$$
