Question

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

Answer 1

`-1/2 ln |1-x|+1/4 ln(1+x^2) +3/2 tan^{-1} x+c`

Explanation:

To break the fraction
`{2-x}/{(1-x)(1+x^2)}`
into partial fractions, we assume

`{2-x}/{(1-x)(1+x^2)} = A/{1-x} + {Bx +C}/{1+x^2}`

Multiplying both sides by `(1-x)(1+x^2)` yields the equation

`2-x = A(1+x^2) +Bx(1-x)+C(1-x)`
`= (A+C) +(B-C) x +(A-B)x^2`

Comparing the coefficients of powers of `x` on both sides gives

`A+C = 2`, `B-C=-1,` and `A-B=0`.

It is easy to see that the solution of this set of equations is `A=B=1/2, C=3/2`, so that

`{2-x}/{(1-x)(1+x^2)} = 1/{2(1-x)} + {1/2x +3/2}/{1+x^2}`

So the required integral is

`int {2-x}/{(1-x)(1+x^2)} dx`
`= int {dx}/{2(1-x)} + 3/2 int {dx}/{1+x^2} + 1/4 int {2xdx}/{1+x^2}`
`= -1/2 ln |1-x|+1/4 ln(1+x^2) +3/2 tan^{-1} x+c`