 # Chris Fox's Engineering Section

## Partial Fractions

Partial fractions consist of two or more fractions, whose numerators consist of a pure number and whose denominators consist of a pure number added to a variable, that can be added together to produce a single fraction whose numerator and denominator both consist of pure numbers and variables.

An example is A/(x+1)+B/(x+2). If A and B both equal 1, the formula becomes 1/(x+1)+1/(x+2). Rearranging this gives {(x+2)+(x+1)}/{(x+1)(x+2)}, and simplifying gives (2x+3)/{(x+1)(x+2)}.

One can also split a single fraction into partial fractions. As an example, (4x+5)/{(x-1)(2x+1)} equates to partial fractions of the form A/(x-1)+B/(2x+1). Rearranging the latter formula gives {A(2x+1)+B(x-1)}/{(x-1)(2x+1)}. Therefore, A(2x+1)+B(x-1)=4x+5. For the values of A and B that we seek, this equation will work for any value of x. So if x=1, the equation becomes 3A+0=9; this leads to A=3. And if x=-0.5, it becomes 0-1.5B=3; this leads to B=-2. Therefore, (4x+5)/{(x-1)(2x+1)}=3/(x-1)-2/(2x+1).

Another partial fraction equation is (2-x²)/{(1+x)(1+x²)}=A/(1+x)+(Bx+C)/(1+x²). This can be rearranged to give 2-x²=A(1+x²)+(Bx+C)(1+x). If x=-1, the equation becomes 1=2A+0; therefore, A=0.5. If x=0, then 2=A+C; it follows that C=1.5. And if x=1, then 1=2A+2B+2C; from this, we can calculate that B=-1.5. The original equation therefore becomes (2-x²)/{(1+x)(1+x²)}=0.5/(1+x)+1.5/(1+x²)-1.5x/(1+x²).