Consider the integral:

$\int \frac{1}{(x+3)(5+2x)}$

My teacher splits this first into two unknown fractions with two unknown numerators, namely:

$\frac{A}{(x+3)}+\frac {B}{(5+2x)}$

He then goes on to perform some sort of magic to find that A = -1 and B = 2

Knowing that my teacher is indeed NOT a magician, I come to you for assistance.

What is my teacher doing and why does it work? Thank you.

  • 3
    $\begingroup$ Partial fractions $\endgroup$ – user61527 Nov 29 '13 at 1:07
  • 9
    $\begingroup$ He's a mathemagician.... $\endgroup$ – Eleven-Eleven Nov 29 '13 at 1:14
  • $\begingroup$ @ChristopherErnst Apparently, my professor had that written on one of his teacher evaluation forms. :) $\endgroup$ – apnorton Nov 29 '13 at 15:45
  • 1
    $\begingroup$ When my students know how to do a step relatively well and I want to skip the step I wave my hands magically and they all sigh at me.... :) $\endgroup$ – Eleven-Eleven Nov 29 '13 at 15:55

If you get a common denominator of $(x+3)(5+2x)$, then the numerators must be equal. Thus, $$1=A(5+2x)+B(x+3)$$ $$1=5A+2Ax+Bx+3B$$ $$0x+1=(2A+B)x+(5A+3B)$$ This means that $$2A+B=0$$ $$5A+3B=1$$ Solve for A and B using substitution or whatever method you prefer. Now you can solve the integral.


Because $\frac{A}{(x+3)}+\frac {B}{(5+2x)} = \frac{A(5+2x)+B(x+3)}{(5+2x)(x+3)} = \frac{(2A+B)x+(5A+3B)}{(5+2x)(x+3)}=\frac{1}{(5+2x)(x+3)}$, this would mean that $2A+B=0$ and $5A+3B=1$. Solving these equations gives $A=-1$ and $B=2$.


You would want to do the following $$ A/(x+3)+B/(5+2x) =1/[(x+3)(5+2x)]$$ $$ A(5+2x) + B(x+3) = 1 $$ Essentially you've multiply both by linear factors of your initial integrand. Equate the coefficients of x and 1.

$$2A+B=0$$ $$5A+3B=1$$

You will find you get the same answer as your teacher.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.