You should notice that the denominator represents the expansion of binomial `(t^2 - 1)^2` such that:

`int (t+4)/(t^4 - 2t^2 + 1)dt = int (t+4)/((t^2 - 1)^2)dt`

You need to use the property of linearity of integrals such that:

`int (t+4)/((t^2 - 1)^2)dt = int t/((t^2 - 1)^2)dt + int 4/((t^2 - 1)^2)dt`

You need to use substitution to evaluate `int t/((t^2 - 1)^2)dt` such that:

`t^2-1 = u => 2tdt = ` du

`int t/((t^2 - 1)^2)dt = int ((du)/2)/(u^2)`

`int ((du)/2)/(u^2) = (1/2)(-1/u) + c`

Substituting back `t^2-1` for u yields:

`int t/((t^2 - 1)^2)dt = -1/2*(1/(t^2-1)) + c`

You need to use partail fraction decomposition to evaluate `int 4/((t^2 - 1)^2)dt ` such that:

`4/((t^2 - 1)^2) = A/(t-1) + B/(t-1)^2 + C/(t+1) + D/(t+1)^2`

`4 = A(t-1)(t+1)^2 + B(t+1)^2 + C(t+1)(t-1)^2 + D(t-1)^2`

`4 = A(t^3 + 2t^2 + t - t^2 - 2t - 1) + B(t^2 + 2t + 1) + C(t^3 - 2t^2 + t + t^2 - 2t + 1) + D(t^2 - 2t + 1)`

`4 = At^3 + t^2(A + B - C + D) + t(-A + 2B - C - 2D) -A + B +C + D`

Equating coefficients of like parts yields:

`A = 0`

`-A + B +C + D = 4 => B +C + D = 4`

`B - C + D = 0`

`2B - C - 2D = 0`

`2B + 2D = 4`

`3B - D = 4 => 3B - D = 2B + 2D => B = 3D`

`6D + 2D = 4 => 8D = 4 => D = 4/8 => D =1/2 => B = 3/2`

`3/2 - C + 1/2 = 0 => 2 - C = 0 => C = 2`

`4/((t^2 - 1)^2) = 3/(2(t-1)^2) + 2/(t+1) + 1/(2(t+1)^2)`

Integrating both sides yields:

`int 4/((t^2 - 1)^2)dt = int 3/(2(t-1)^2)dt + int 2/(t+1)dt + int 1/(2(t+1)^2)dt`

`int 4/((t^2 - 1)^2)dt = -3/(2(t-1)) + 2ln|t+1| - 1/(2(t+1)) + c`

**Hence, evaluating the given integral yields `int (t+4)/((t^2 - 1)^2)dt = -1/2*(1/(t^2-1))- 3/(2(t-1)) + 2ln|t+1| - 1/(2(t+1)) + c.` **

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