Question

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

Answer 1

`int 1/(n(n+2)) dn = 1/2 ln abs(n) - 1/2 ln abs(n+2) + C`

Explanation:

Since we have already been given a factorisation of the denominator into distinct linear factors, we are looking for a partical fraction decomposition of the form:

`1/(n(n+2)) = A/n + B/(n+2)`

`=(A(n+2)+Bn)/(n(n+2))`

`=((A+B)n+2A)/(n(n+2))`

Equating coefficients we find:

`{(A+B=0), (2A=1) :}`

Hence:

`{(A=1/2),(B=-1/2):}`

So:

`int 1/(n(n+2)) dn = int (1/(2n) - 1/(2(n+2))) dn = 1/2 ln abs(n) - 1/2 ln abs(n+2) + C`