Question

If $\int \frac{8}{(4n-1)(4n+3)}dx$ .find integration of this function.

Answer 1

$\int \frac{B}{(4n-1)(4n+3)}dn=\int B[\frac{A}{(4n-1)}+\frac{B}{(4n+3)}]dn$

$\Rightarrow \frac{1}{(4n-1)(4n+3)}=\frac{A}{(4n-1)}+\frac{B}{(4n+3)}$

$=\frac{4An+3A+4Bn-B}{(4n-1)(4n+3)}$

$\frac{(4A+4B)n+(BA-B)}{(4n-1)(4n+3)}$

compare both side

$1=(4A+4B)n+(3A-B)\Rightarrow 4A+4B=0$......$\left(1\right)$

$3A-B=1$.....$\left(2\right)$

$4A=-4B\Rightarrow A=-B$......$\left(3\right)$

By $\left(2\right)$ & $\left(3\right)$ $4A=1\Rightarrow a=\frac{1}{4}b=-\frac{1}{4}$

So: $\int \frac{B}{(4n-1)(4n+3)}dn=\int \frac{B}{4}\left[\frac{1}{(4n-1)-\frac{1}{(4n-1)}}\right]dn=\frac{2\log (4n-1)}{4}-\frac{2\log (4n+3)}{4}$

$=\frac{1}{2}\left[\log \left(\frac{4n-1}{4n+3}\right)\right]$