Question

Find `int \ (x^2+x+1)/(x^2(x+2)) \ dx` using partial fractions?

Answer 1

`int \ (x^2+x+1)/(x^2(x+2)) \ dx = 1/4ln|x| - 1/(2x)+ 3/4 ln |x+2|+C`

Explanation:

We seek:

`I = int \ (x^2+x+1)/(x^2(x+2)) \ dx`

We can decompose the integrand into partial fractions:

`(x^2+x+1)/(x^2(x+2)) -= A/x + B/x^2 + C/(x+2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( Ax(x+2) + B(x+2) + Cx^2 ) / (x^2(x+2))`

Leading to:

`x^2+x+1 -= Ax(x+2) + B(x+2) + Cx^2`

Where `A,B,C` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x = 0 => 1= 2B => B =1/2`
Put `x = -2 => 3 = 4C => C=3/4`
Coeff`(x^2): 1 = A+C => A = 1/4`

Hence, we can now write the integral as:

`I = int \ (1/4)/x + (1/2)/x^2 + (3/4)/(x+2) \ dx`
`\ \ = 1/4 int \ 1/x \ dx + 1/2int \ 1/x^2 \ dx + 3/4 \ int 1/(x+2) \ dx`
`\ \ = 1/4ln|x| +1/2 x^(-1)/(-1) + 3/4 ln |x+2|+C`
`\ \ = 1/4ln|x| - 1/(2x)+ 3/4 ln |x+2|+C`