Question

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

Answer 1

`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 !

Answer 2

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`