How do you use partial fraction decomposition to decompose the fraction to integrate 4/((x^2+9)(x+1))?

2 Answers
May 19, 2018

int4/((x²+9)(x+1))dx=-1/5ln(x²+9)+2/15arctan(x/3)+2/5ln(|x+1|)+ C
C in RR

Explanation:

int4/((x²+9)(x+1))dx=4int1/((x²+9)(x+1))dx
let 1/((x²+9)(x+1))=(Ax+B)/(x²+9)+C/(x+1)
1=(Ax+B)(x+1)+C(x²+9)
1=Ax²+Ax+Bx+B+Cx²+9C

1=(A+C)x²+(A+B)x+9C
By identification :

A+C=0[1]
A+B=0[2]
9C+B=1[3]

A+C=0
B-C=0
9C+B=1[3]

A+C=0
B-C=0
10C=1

A=-1/10
B=1/10
C=1/10

So: 4int1/((x²+9)(x+1))dx=4int((-1/10x+1/10)/(x²+9)+(1/10)/(x+1))dx

=-4/10intx/(x²+9)dx+4/10int1/(x²+9)dx+4/10int1/(x+1)dx
=-2/10int(2x)/(x²+9)dx+2/5int1/(x²+3²)dx+2/5int1/(x+1)dx
Because int(u')/udu=ln(|u|), int1/(x²+b²)dx=1/b*arctan(x/b),
we have :

int4/((x²+9)(x+1))dx=-1/5ln(x²+9)+2/15arctan(x/3)+2/5ln(|x+1|)+ C
C in RR

\0/ Here's our answer !

May 19, 2018

See the explanation below

Explanation:

The decomposition into partial fractions is

4/((x^2+9)(x+1))=(Ax+B)/(x^2+9)+C/(x+1)

=((Ax+B)(x+1)+C(x^2+9))/((x^2+9)(x+1))

The denominators are the same, compare the numerators

4=(Ax+B)(x+1)+C(x^2+9)

Let x=-1, =>, 4=10C, =>, C=2/5

The coefficients of x^2 are

0=A+C, =>, A=-C=-2/5

Let x=0

4=B+9C, =>, B=4-9C=4-18/5=2/5

Therefore,

4/((x^2+9)(x+1))=(-2/5x+2/5)/(x^2+9)+(2/5)/(x+1)

So, the integral is

I=int(4dx)/((x^2+9)(x+1))=int((-2/5x+2/5)dx)/(x^2+9)+int(2/5dx)/(x+1)

=2/5ln(|x+1|)+I_1

I_1=2/5int((-x+1)dx)/(x^2+9)

=2/5int(-xdx)/(x^2+9)+2/5int(dx)/(x^2+9)

=-2/5*1/2ln(x^2+9)+2/5*1/9*3arctan(x/3)

Finally,

int(4dx)/((x^2+9)(x+1))=2/5ln(|x+1|)-1/5ln(x^2+9)+2/15arctan(x/3)+C