Question

Add

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

Answer 1

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)`