How do you integrate #int (2x+1)/(4x^2+12x-7) dx# using partial fractions?

1 Answer
Narad T.
Oct 26, 2016

The integral #int((2x+1)dx)/(4x^2+12x-7)=3/8ln(2x+7)+1/8ln(2x-1)+C#

Explanation:

Let's factorise the denominator
#4x^2+12x-7=(2x+7)(2x-1)#

We look for the partial fractions
#(2x+1)/(4x^2+12x-7)=A/(2x+7)+B/(2x-1)#

#(2x+1)/(4x^2+12x-7)=(A(2x-1)+B(2x+7))/(4x^2+12x-7)#

so, #2x+1=A(2x-1)+B(2x+7))#
Let #x=1/2# #=>##2=8B# #=>##B=1/4#
Let #x=0##=>##1=-A+7/4##=>##A=3/4#

#(2x+1)/(4x^2+12x-7)=(3/4)/(2x+7)+(1/4)/(2x-1)#
Let's do the integral
#int((2x+1)dx)/(4x^2+12x-7)=int((3/4)dx)/(2x+7)+int((1/4)dx)/(2x-1)#

#=3/8ln(2x+7)+1/8ln(2x-1)+C#

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