Partial Fraction

What are Partial Fraction?

As we can know we can do this
                                            . $$\frac{9}{x+4}+\frac{2}{x-1}=\frac{11x-1}{\left(x+4\right)\left(x-1\right)}$$
like this
$$\begin{array}{l}\frac{9}{x+4}+\frac{2}{x-1}=\frac{9(x+1)+2(x+4)}{\left(x+4\right)(x-1)}\\ =\frac{9x-9+2x+8}{\left(x+4\right)\left(x-1\right)}\\ =\frac{11x-1}{\left(x+4\right)\left(x-1\right)}\end{array}$$

but how can we do the reverse?

That process is known as Partial Fraction.

So now let we see how to do this.

        

Step 1    factor the denominator

in our example denominator is already in factorized form

Step 2  Take  two constants A & B (because in denominator we have two factor i.e. (x + 4), (x - 1) ).

$$\frac{11x-1}{\left(x+4\right)\left(x-1\right)}=\frac{A}{\left(x+4\right)}+\frac{B}{\left(x-1\right)}$$

Step 3  Multiply through by the bottom so we no longer have fractions
$$11x-1=A\left(x-1\right)+B\left(x+4\right)$$

Step 4  Now find the value of constants( i.e. A & B)
                 Substituting the roots, or "zeros", of  (x + 4) & (x - 1)
$$\begin{array}{l}forx-1=0\Rightarrow x=1\\ 11\left(1\right)-1=A\left(\text{1-1}\right)+B\left(1+4\right)\\ 10=A\left(0\right)+B\left(5\right)\\ B=2\\ Againx+4=0\Rightarrow x=-4\\ 11\left(-4\right)-1=A\left(\text{-4-1}\right)+B\left(-4+4\right)\\ -45=A\left(-5\right)+B\left(0\right)\\ A=9\end{array}$$

NOW THIS IS OUR ANSWER $$\frac{11x-1}{\left(x-1\right)\left(x+4\right)}=\frac{9}{\left(x+4\right)}+\frac{2}{\left(x-1\right)}$$