How do you integrate #4/((x+1)(x-5))# using partial fractions?

2 Answers
Feb 18, 2017

The answer is #=-2/3ln(|x+1|)+2/3ln(|x-5|)+C#

Explanation:

Let's perform the decomposition into partial fractions

#4/((x+1)(x-5))=A/(x+1)+B/(x-5)#

#=(A(x-5)+B(x+1))/((x+1)(x-5))#

As the denominators are the same, we compare the numerators

#4=A(x-5)+B(x+1)#

Let #x=-1#, #=>#, #4=-6A#, #=>#, #A=-2/3#

Let #x=5#, #=>#, #4=6B#, #=>#, #B=2/3#

Therefore,

#4/((x+1)(x-5))=(-2/3)/(x+1)+(2/3)/(x-5)#

So,

#int(4dx)/((x+1)(x-5))=-2/3intdx/(x+1)+2/3intdx/(x-5)#

#=-2/3ln(|x+1|)+2/3ln(|x-5|)+C#

#=2/3ln((|x-5|)/(|x+1|))+C#

Feb 18, 2017

#int 4/((x+1)(x-5)) dx=color(green)(2/3(ln((abs(x-5))/(abs(x+1))) +C)#

Explanation:

Step 1: Partial Fraction Decomposition
If #4/((x+1)(x-5))=A/(x+1)+B/(x-5)#
then
#color(white)("XXX")A(x-5)+B(x+1)=4#

#color(white)("XXX")rarr Ax+Bx = 0x color(white)("XX")rarr A+B=0color(white)("XX")rarr A=-B#
and
#color(white)("XXX")rarr -5A+B=4#
Combining:
#color(white)("XXX")-5(-B)+B=4color(white)("XX")B=2/3color(white)("XX")rarr A=-2/3#

#4/((x+1)(x-5))=-2/3(1/(x+1))+2/3(1/(x-5))#

Part 2: Integration Using the Fractional Decompostion
#int 4/((x+1)(x-5)) dx#

#color(white)("XXX")=intcolor(white)("x")[-2/3(1/(x+1))+2/3(1/(x-5))]dx#

#color(white)("XXX")=-2/3int1/(x+1)dx+2/3int1/(x-5) dx#

#color(white)("XXXXXXXX")#Somewhere around here, it would be convenient to remember:
#color(white)("XXXXXXXXXXX")int 1/(x+a) dx = lnabs(x+a) +C#
#color(white)("XXXXXXXXXXX")#so we can continue...

#color(white)("XXX")=-2/3lnabs(x+1)+2/3lnabs(x-5)+C#

#color(white)("XXX")=2/3[lnabs(x-5)-lnabs(x+1)]+C#

#color(white)("XXX")=2/3ln((abs(x-5))/(abs(x+1))) +C#