Question

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

Answer 1

`int(2x+1)/((x-2)(x^2+1))dx=ln|x-2|-1/2ln(x^2+1)+c`

Explanation:

`int(2x+1)/((x-2)(x^2+1))dx=intA/(x-2)dx+int(Bx+C)/(x^2+1)dx`
`0x^2+2x+1=A(x^2+1)+(Bx+C)(x-2)`
Equating coefficients:
`A+B=0`
`C-2B=2`
`A-2C=1`
Solving simultaneously yields `A=1`, `B=-1` and `C=0`
`=int1/(x-2)dx-intx/(x^2+1)dx`
`=ln|x-2|-1/2ln(x^2+1)+c`