Help with $\int \frac{1}{(1-t^2)t^2} \, dt$ $$\int \frac{1}{(1-t^2)t^2} \, dt$$
By using partial fractions I get:
$$\frac{1}{(1-t)(1+t)t^2} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t+1} + \frac{D}{1-t}$$
$$1 = t^3 (-A-C+D) + t^2 (-B+C+D) + At+ B$$
$$-A-C+D = 0$$
$$-B+C+D=0$$
$$A=0$$
$$B=1$$
So, $A=0, B=1, C=\frac{1}{2}, D = \frac{1}{2}$
Then by replacing the variables with the results we get:
$$\int \frac{1}{(1-t^2)t^2} \, dt = \int \frac{1}{t^2} \, dt + \frac{1}{2} \int \frac{1}{t+1} \, dt + \frac{1}{2} \int \frac{1}{1-t} \, dt$$
$$- \frac{1}{t} + C + \frac{1}{2} \ln|t+1| + C + \frac{1}{2} \ln|1-t| + C$$
$$- \frac{1}{t} + \frac{1}{2} \ln|t+1| + \frac{1}{2} \ln|1-t| + C$$
Is this correct?

We may utilize $$\dfrac1{t^2(1-t^2)}=\dfrac{(1-t^2)+t^2}{t^2(1-t^2)}=\dfrac1{t^2}+\dfrac1{1-t^2}$$
Finally $$\dfrac1{1-t^2}=\dfrac{1+t+1-t}{2(1-t)(1+t)}=?$$
