How do you integrate #int (1-x^2)/((x+1)(x-6)(x+2)) # using partial fractions?

Question

1004 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-13T06:57:50

THe answer is #=-5/8ln(|x-6|)-3/8ln(|x+2|)+C#

Perform the decomposition into partial fractions

#(1-x^2)/((x+1)(x-6)(x+2))=A/(x+1)+B/(x-6)+C/(x+2)#

#=(A(x-6)(x+2)+B(x+1)(x+2)+C(x+1)(x-6))/((x+1)(x-6)(x+2))#

The denominators are the same, compare the numerators

#1-x^2=A(x-6)(x+2)+B(x+1)(x+2)+C(x+1)(x-6)#

Let #x=-1#, #=>#, #0=A#

Let #x=6#, #=>#, #-35=B*7*8#, #=>#, #B=-5/8#

Let #x=-2#, #=>#, #-3=C*-1*-8#, #C=-3/8#

Therefore,

#(1-x^2)/((x+1)(x-6)(x+2))=(0)/(x+1)+(-5/8)/(x-6)+(-3/8)/(x+2)#

So,

#int((1-x^2)dx)/((x+1)(x-6)(x+2))=int(-5/8dx)/(x-6)+int(-3/8dx)/(x+2)#

#=-5/8ln(|x-6|)-3/8ln(|x+2|)+C#

Answer 2 · 2018-01-13T19:07:31

#int (1-x^2)/[(x+1)(x-6)(x+2)]*dx#

=#-3/8Ln(x+2)-5/8Ln(x-6)+C#

#int (1-x^2)/[(x+1)(x-6)(x+2)]*dx#

=-#int (x^2-1)/[(x+1)(x-6)(x+2)]*dx#

=-#int ((x+1)(x-1))/[(x+1)(x-6)(x+2)]*dx#

=-#int (x-1)/[(x-6)*(x+2)]*dx#

=#int (1-x)/[(x-6)*(x+2)]*dx#

Now I decomposed integrand,

#(1-x)/[(x-6)*(x+2)]=A/(x+2)+B/(x-6)#

After expanding denominator,

#A*(x-6)+B*(x+2)=1-x#

Set #x=-2#, #-8A=3#. So #A=-3/8#

Set #x=6#, #8B=-5#. So #B=-5/8#

Hence,

#int (1-x)/[(x-6)*(x+2)]*dx#

=#-3/8 int (dx)/(x+2)-5/8 int (dx)/(x-6)#

=#-3/8Ln(x+2)-5/8Ln(x-6)+C#