How do you 1/(1-t^2) in partial fractions?

You have entered [src]

  1   
------
     2
1 - t

$$\frac{1}{1 - t^{2}}$$

1/(1 - t^2)

Fraction decomposition [src]

1/(2*(1 + t)) - 1/(2*(-1 + t))

$$\frac{1}{2 \left(t + 1\right)} - \frac{1}{2 \left(t - 1\right)}$$

    1           1     
--------- - ----------
2*(1 + t)   2*(-1 + t)

General simplification [src]

  -1   
-------
      2
-1 + t

$$- \frac{1}{t^{2} - 1}$$

-1/(-1 + t^2)

Common denominator [src]

  -1   
-------
      2
-1 + t

$$- \frac{1}{t^{2} - 1}$$

-1/(-1 + t^2)

Numerical answer [src]

1/(1.0 - t^2)

1/(1.0 - t^2)

Combinatorics [src]

      -1        
----------------
(1 + t)*(-1 + t)

$$- \frac{1}{\left(t - 1\right) \left(t + 1\right)}$$

-1/((1 + t)*(-1 + t))