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

1 Answer
Apr 14, 2016

$\int \frac{1}{n \left(n + 2\right)} \mathrm{dn} = \frac{1}{2} \ln \left\mid n \right\mid - \frac{1}{2} \ln \left\mid n + 2 \right\mid + C$

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:

$\frac{1}{n \left(n + 2\right)} = \frac{A}{n} + \frac{B}{n + 2}$

$= \frac{A \left(n + 2\right) + B n}{n \left(n + 2\right)}$

$= \frac{\left(A + B\right) n + 2 A}{n \left(n + 2\right)}$

Equating coefficients we find:

$\left\{\begin{matrix}A + B = 0 \\ 2 A = 1\end{matrix}\right.$

Hence:

$\left\{\begin{matrix}A = \frac{1}{2} \\ B = - \frac{1}{2}\end{matrix}\right.$

So:

$\int \frac{1}{n \left(n + 2\right)} \mathrm{dn} = \int \left(\frac{1}{2 n} - \frac{1}{2 \left(n + 2\right)}\right) \mathrm{dn} = \frac{1}{2} \ln \left\mid n \right\mid - \frac{1}{2} \ln \left\mid n + 2 \right\mid + C$