Question

How do you integrate `7/(x^2+1)` using partial fractions?

Answer 1

`int 7/(x^2+1) dx = 7/2 ln(x+i) - 7/2 ln(x-i) + C`

Explanation:

This integral would normally be done using trigonometric substitution, but if you really want to use partial fractions to integrate this then you will need to use Complex coefficients:

`7/(x^2+1) = A/(x+i)+B/(x-i) = (A(x-i)+B(x+i))/(x^2+1)`

`=((A+B)x+(B-A)i)/(x^2+1)`

Equating coefficients:

`{ (A+B=0), ((B-A)i = 7) :}`

Multiply both sides of the second equation by `i` to get:

`A-B=7i`

Hence `A=7/2i` and `B=-7/2i`
So:

`int 7/(x^2+1) dx = int 7/(2(x+i))-7/(2(x-i)) dx`

`= 7/2 ln(x+i) - 7/2 ln(x-i) + C`