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

2 Answers
Narad T.
Jan 13, 2018

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

Explanation:

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

Cem Sentin
Jan 13, 2018

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

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

Explanation:

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

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