# simplify the equation [1/(x-1)-1/(x+1)+1]*(x+1)/(x^2+1)

### 2 Answers | Add Yours

We have to simplify : [1/(x-1)-1/(x+1)+1]*(x+1)/(x^2+1)

[1/(x-1)-1/(x+1)+1]*(x+1)/(x^2+1)

=> [((x+1) - (x - 1) + (x^2 - 1))/(x^2 -1)]*(x+1)/(x^2+1)

=> [(x + 1 - x + 1 + x^2 - 1)/(x^2 -1)]*(x+1)/(x^2+1)

=> [(x^2 + 1)/(x^2 -1)]*(x+1)/(x^2+1)

=> (x^2 + 1)*(x+1)/(x^2 -1)*(x^2+1)

=> (x+1)/(x^2 -1)

=> (x+1)/(x - 1)(x + 1)

=> 1/(x - 1)

**The simplified form is 1/ (x - 1)**

We'll multiply by LCD inside the first pair of brackets:

LCD = (x-1)(x+1) = x^2 - 1

[(x+1-x+1+x^2-1)/(x^2-1)]*[(x+1)/(x^2+1)]

We'll combine and eliminate like terms:

[(x^2 + 1)/(x^2-1)]*[(x+1)/(x^2+1)]

We'll simplify by (x^2+1):

[(x^2 + 1)/(x^2-1)]*[(x+1)/(x^2+1)] = (x+1)/(x^2-1)

W'll re-write the difference of squares from denominator as a product:

(x+1)/(x^2-1) = (x+1)/(x-1)(x+1)

We'll simplify by (x+1):

E(x) = 1/(x-1)

**The simplified expression is E(x) = 1/(x-1).**