March 19th, 2017, 10:18 AM  #1 
Newbie Joined: Mar 2017 From: Budapest Posts: 1 Thanks: 0  Improper integral
I attached the problem below. My question is where does the 1/4 come from? I see that 1/((x+3)(x1)) becomes (1/(x1))(1/(x+3)) because of the 1/4 we took out, but I don't understand this step. Would someone please clarify? Thank you!

March 19th, 2017, 10:30 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,569 Thanks: 1272 
$\dfrac{1}{(x+3)(x1)} = \dfrac{A}{x+3}+\dfrac{B}{x1}$ $1 = A(x1) + B(x+3)$ $x=1 \implies 1 = 4B \implies B = \dfrac{1}{4}$ $x=3 \implies 1 = 4A \implies A = \dfrac{1}{4}$ $\dfrac{1}{(x+3)(x1)} = \dfrac{1}{4}\bigg[\dfrac{1}{x1}\dfrac{1}{x+3}\bigg]$ 

Tags 
improper, improper integral, integral, integral calculus 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
improper integral  alexmath  Calculus  1  April 1st, 2014 08:47 AM 
Improper integral  jones1234  Calculus  1  December 23rd, 2012 08:51 AM 
ODE & Improper Integral  talisman  Real Analysis  13  December 5th, 2011 02:42 PM 
improper integral  izseekzu  Calculus  1  April 13th, 2010 03:37 PM 
Improper integral  tnutty  Calculus  2  February 23rd, 2009 02:12 PM 