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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.

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    $\begingroup$ Partial fractions $\endgroup$ – user61527 Nov 29 '13 at 1:07
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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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 $B=2$.

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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.

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