How do you combine (x-1)*[3/(x^2-1)+x/(2x-2)]?

Aug 22, 2017

$= \frac{{x}^{2} + x + 6}{2 \left(x + 1\right)}$

Explanation:

$\left(x - 1\right) \cdot \left[\frac{3}{{x}^{2} - 1} + \frac{x}{2 x - 2}\right]$

Factorise the denominators first

$= \left(x - 1\right) \cdot \left[\frac{3}{\left(x + 1\right) \left(x - 1\right)} + \frac{x}{2 \left(x - 1\right)}\right]$

Add the fractions using a common denominator.

=(x-1)*[((2xx3 + x(x+1))/(2(x+1)(x-1))]

$= \left(x - 1\right) \cdot \left[\frac{6 + {x}^{2} + x}{2 \left(x + 1\right) \left(x - 1\right)}\right]$

Distribute $\left(x - 1\right)$ into the bracket:

=[(cancel((x-1))(6 + x^2+x))/(2(x+1)cancel((x-1))]

$= \frac{{x}^{2} + x + 6}{2 \left(x + 1\right)}$