How do you integrate #int (5x)/((x-1)(x+3))# using partial fractions?

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Answer 1 · 2016-11-11T16:34:13

The answer is #=5/4ln(x-1)+15/4ln(x+3)+C#

First, let's do the decomposition into partial fractions,
#(5x)/((x-1)(x+3))=A/(x-1)+B/(x+3)#

#=((x+3)+B(x-1))/((x-1)(x+3))#

#:. 5x=A(x+3)+B(x-1)#
Let #x=-3##=>##-15=-4B##=>##B=15/4#

Let #x=1##=>##5=4A##=>##A=5/4#

#(5x)/((x-1)(x+3))=(5/4)/(x-1)+(15/4)/(x+3)#

So, #int(5xdx)/((x-1)(x+3))=int(5/4dx)/(x-1)+int(15/4dx)/(x+3)#

#=5/4ln(x-1)+15/4ln(x+3)+C#