How do you integrate #int (4x^3 + 12x^2 - x - 4 )/( 4x^2 - 1)# using partial fractions?

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Answer 1 · 2016-12-12T06:47:02

The answer is #=x^2/2+3x+1/4ln(∣2x+1∣)-1/4ln(∣2x-1∣)+C#

We use #a^2-b^2=(a+b)(a-b)#

#intx^ndx=x^(n+1)/(n+1)# and #intdx/x=∣x∣#

Let's do a long division

#color(white)(aaaa)##4x^3-x+12x^2-4##color(white)(aaaa)##∣##4x^2-1#

#color(white)(aaaa)##4x^3-x##color(white)(aaaaaaaaaaaaaa)##∣##x+3#

#color(white)(aaaaaa)##0 -0+12x^2-4#

#color(white)(aaaaaaaaaaa)##+12x^2-3#

#color(white)(aaaaaaaaaaaaaa)##+0-1#

So,

#(4x^3-x+12x^2-4)/(4x^2-1)=x+3-1/(4x^2-1)#

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

Let's do the decomposition into partial fractions

#1/((2x+1)(2x-1))=A/(2x+1)+B/(2x-1)#

#=(A(2x-1)+B(2x+1))/((2x+1)(2x-1))#

Therefore,

#1=A(2x-1)+B(2x+1)#

Let #x=1/2#,#=>#,#1=2B#, #=>#, #B=1/2#

Let #x=-1/2#,#=>#,#1=-2A#, #=>#, #A=-1/2#

So,

#(4x^3-x+12x^2-4)/(4x^2-1)=x+3-((-1/2)/(2x+1)+(1/2)/(2x-1))#

#int((4x^3-x+12x^2-4)dx)/(4x^2-1)=int(x+3)dx+1/2int(dx)/(2x+1)-1/2int(dx)/(2x-1)#

#=x^2/2+3x+1/4ln(∣2x+1∣)-1/4ln(∣2x-1∣)+C#