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(k+5)/(k+1) + (8)/(k^2-1)

User Marsu
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Get a common denominator which you can get by mutiplying by k-1

((k + 5)(k - 1))/((k + 1)(k - 1)) + \frac{8}{ {k}^(2) - 1}
Now foil and distribute

\frac{ {k}^(2) + 4k - 5}{ {k}^(2) - 1} + \frac{8}{ {k}^(2) - 1 }
Now combine the fractions and like terms

\frac{ {k}^(2) + 4k - 5 + 8}{ {k}^(2) - 1 } = \frac{ {k}^(2) + 4k + 3}{ {k}^(2) - 1 }
Now factor out a k+1

((k + 1)(k + 3))/((k + 1)(k - 1))
you can cancel out the k+1 leaving

(k + 3)/(k - 1)
User Nicolas Lauquin
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