Sure, here's how to approach this question:
Let's first factorise the denominator of the given expression:
x²+1 can't be factorised any further, so let's work on the second factor:
x²+3x-2 = (x-1)(x+2)
Therefore, the denominator of the given expression becomes: (x²+1)(x-1)(x+2)
Now, let's assume that the expression is equal to a sum of two fractions:
1-2x²/(x²+1)(x-1)(x+2) = A/(x²+1) + B/(x-1) + C/(x+2)
Multiplying both sides by the denominator, we get:
1-2x² = A(x-1)(x+2) + B(x²+1)(x+2) + C(x²+1)(x-1)
Let's first put x=1 and solve for B:
1-2(1)² = B(1²+1)(1+2) + C(1²+1)(1-1)
1-2 = 4B
B = -1/2
Next, let's put x=-2 and solve for C:
1-2(-2)² = A(-2-1)(-2+2) + B((-2)²+1)((-2)+2) + C((-2)²+1)(-2-1)
1-8 = 15C
C = 7/-15
Finally, we can solve for A by putting x=0:
1-2(0)² = A(0-1)(0+2) + B((0)²+1)((0)+2) + C((0)²+1)(0-1)
1-0 = -2A + 2B - C
1 = -2A + 2(-1/2) - (7/15)
6/15 = -2A
A = -3/10
Therefore, the given expression can be resolved into partial fractions as:
1-2x²/(x²+1)(x-1)(x+2) = (-3/10)/(x²+1) + (-1/2)/(x-1) + (7/15)/(x+2)
I hope that helps!