Let $f(x) = \frac{-2x+4}{(x^2+1)(x-1)^2}$. Express the function $f(x)$ as a sum of partial fractions.

Question

I have gotten

$(-2x+4) = A(x-1)^2 + B((x^2+1))$

But after letting $x = 1$, $B = 1$,

I couldn't find $A$.

Any help is appreciated.

Thank you

Answer 1

Because there is a quadratic factor and two linear factors in the denominator, you can write the fraction in the form:

$\frac{-2x+4}{(x^2+1)(x-1)^2}=\frac{Ax+B}{(x^2+1)}+\frac{C}{(x-1)}+\frac{D}{(x-1)^2}$

$-2x+4=(Ax+B)(x-1)^2+C(x^2+1)(x-1)+D(x^2+1)$

$-2x+4=(Ax+B)(x-1)^2+C(x^2+1)(x-1)+Dx^2+D$

$-2x+4=(Ax+B)(x^2-2x+1)+C(x^3-x^2+x-1)+Dx^2+D$

$-2x+4={\color{red}{Ax^3}}\color{blue}{-2Ax^2}+\color{green}{Ax}+\color{blue}{Bx^2}\color{green}{-2Bx}+B+\color{red}{Cx^3}\color{blue}{-Cx^2}+\color{green}{Cx}-C+\color{blue}{Dx^2}+D$

$-2x+4=\color{red}{x^3(A+C)}+\color{blue}{x^2(-2A+B-C+D)}+\color{green}{x(A-2B+C)}+(B-C+D)$

From that, we can get the following equations:$$\begin{align}&(1)A+C=0\\&(2)-2A+B-C+D=0\\&(3)A-2B+C=-2\\&(4)B-C+D=4\end{align}$$Plugging in $(1)$ into $(3)$:

$-2B=-2$

$B=1$

Subtracting $(4)$ from $(2)$:

$-2A+B-C+D-(B-C+D)=0-4$

$-2A=-4$

$A=2$

Plug that into $(1)$ to get $C=-2$

Plug $A=2,B=1,C=-2$ into $(2)$:

$-2(2)+1-(-2)+D=0$

$-4+1+2+D=0$

$D-1=0\implies D=1$

So we have $A=2,B=1,C=-2,D=1$

The final answer is:$$\frac{-2x+4}{(x^2+1)(x-1)^2}=\boxed{\frac{2x+1}{(x^2+1)}+\frac{-2}{(x-1)}+\frac{1}{(x-1)^2}}$$

Test:

$\frac{2x+1}{(x^2+1)}+\frac{-2}{(x-1)}+\frac{1}{(x-1)^2}=\frac{(2x+1)(x-1)^2}{(x^2+1)(x-1)^2}+\frac{-2(x-1)(x^2+1)}{(x-1)^2(x^2+1)}+\frac{x^2+1}{(x^2+1)(x-1)^2}=\frac{(2x+1)(x-1)^2-2(x-1)(x^2+1)+x^2+1}{(x^2+1)(x-1)^2}=\frac{(2x+1)(x^2-2x+1)-2(x^3-x^2+x-1)+x^2+1}{(x^2+1)(x-1)^2}=\frac{(2x^3-3x^2+1)+(-2x^3+2x^2-2x+2)+(x^2+1)}{(x^2+1)(x-1)^2}=\frac{(2x^3-2x^3)+(-3x^2+2x^2+x^2)+(-2x)+(1+2+1)}{(x^2+1)(x-1)^2}=\frac{-2x+4}{(x^2+1)(x-1)^2}$

Answer 2

I think you finish half of the work. Then you just let $x=0$, $$4=A+B$$ Plug $B=1$, get $A=3$

Answer 3

Your approach is not correct.

You need to start with

$$\frac{-2x+4}{(x^2+1)(x-1)^2}= \frac{Ax+B}{x^2+1} + \frac C{x-1} + \frac D{(x-1)^2}$$

Hence,

$$-2x+4 = (Ax+B)(x-1)^2+C(x-1)(x^2+1) + D(x^2+1)$$

This will lead to $$\frac{-2x+4}{(x^2+1)(x-1)^2}= \frac{2x+1}{x^2+1} - \frac 2{x-1} + \frac 1{(x-1)^2}$$