Question

Please help me do this question

Answer 1

Step-by-step explanation:

We need to break up the given expression into two separate fractions so that when they are added together, we will get the original expression.

Note that the denominator is made up of two factors, $(2x - 1)$ and $(x^2 + 1).$ But note that the 2nd factor is a 2nd order polynomial so as a rule, the numerator of the fraction containing this factor must be an (n - 1)-order polynomial, where n is the order. With that in mind, we can write the general form of the partial fractions as follows:

$\dfrac{x + 7}{(2x - 1)(x^2 + 1)} = \dfrac{Ax + B}{x^2 + 1} + \dfrac{C}{2x - 1}$

$\:\:\:\:=\dfrac{2Ax^2 + Bx - Ax - B + Cx^2 + C}{(2x - 1)(x^2 + 1)}$

Here, we combined the two fractions to form the equation above. Next, we compare the numerators on either side. Note that we satisfy the equality if we impose the following conditions:

$2A + C = 0\:\:\:\:\:\:\:\:\:\:\:\:\:\:(1)$

$2B - \:A = 1\:\:\:\:\:\:\:\:\:\:\:\:\:\:(2)$

$-B + C = 7\:\:\:\:\:\:\:\:\:\:\:\:\:\:(3)$

To find the values of the constants A, B and C, we need to solve this system of equations. We start by multiplying Eqn(3) by 2 and then adding it to Eqn(2) to get

$-A + 2C = 15\:\:\:\:\:\:\:\:\:\:\:\:\:\:(4)$

Then, multiply Eqn(1) by -2 to get

$-4A - 2C = 0\:\:\:\:\:\:\:\:\:\:\:\:\:\:(5)$

Add Eqn(4) to Eqn(5) and we will get

$-5A = 15 \Rightarrow A = -3$

Now that we know the value of A, we can use this in Eqn(2) to get

$2B - (-3) = 1 \Rightarrow B = -1$

Next, to solve for C, we use the value of A in Eqn(1) to get

$2(-3) + C = 0 \Rightarrow C = 6$

Therefore, the given expression can be written as

$\dfrac{x + 7}{(2x - 1)(x^2 + 1)} = \dfrac{6}{2x - 1} + \left(\dfrac{-3x - 1}{x^2 + 1}\right)$

As a check, let's combine the two fractions together:

$\dfrac{x + 7}{(2x - 1)(x^2 + 1)} = \dfrac{6}{2x - 1} + \left(\dfrac{-3x - 1}{x^2 + 1}\right)$

$= \dfrac{6x^2 + 6 - 6x^2 - 2x + 3x + 1}{(2x - 1)(x^2 + 1)}$

$= \dfrac{x + 7}{(2x - 1)(x^2 + 1)}$

As expected, we got the original expression.