Express 3+5/(x+1)(2x^2+7) as a sum of partial fractions

Answer 1

To express the given expression as a sum of partial fractions, we first decompose the denominator:

3+5/(x+1)(2x^2+7) = A/(x+1) + Bx + C/(2x^2+7)

Now, we need to find the values of A, B, and C. We first clear the fractions:

3 + 5/(x+1)(2x^2+7) = A(2x^2+7) + (Bx + C)(x+1)

Now, let's expand the expression:

5 = A(2x^2+7) + (Bx + C)(x+1) - 3(x+1)(2x^2+7).

Choose appropriate values for x that simplify the equation and solve for A, B, and C:

For x = -1,

5 = A(2(-1)^2 + 7) -> 5 = 9A -> A = 5/9.

For x = 0,

5 = (B(0) + C)(0 + 1) -> 5 = C -> C = 5.

For simplicity, let's use A and C in the expanded equation to solve for B:

5 = (5/9)(2x^2+7) + (Bx + 5)(x+1) - 3(x+1)(2x^2+7)

Now, let's collect the terms of the same power of x:

x^2 terms: 0 = -5/9(2) + B + 3(2)
x terms: 0 = -5 + B
constant term: 5 = 5/9(7) + 5 - 3*7

Solve for B:

0 = B - 5 -> B = 5.

Now we can write the given expression as a sum of partial fractions:

3+5/(x+1)(2x^2+7) = (5/9)/(x+1) + 5x/ (2x^2+7) + 5/(2x^2+7).