Question

Evaluate:$\int \frac{dt}{(1-t^2)(1-2t^2)}$

Answer 1

$\frac{1}{(1-t^2)(1-2t^2)}=\frac{1}{(1-2t^2)}-\frac{1}{2(1-t^2)}$

$\frac{1}{(1-2t^2)}=\frac{1}{(1-\sqrt{2}t)(1+\sqrt{2}t)}=\frac{1}{2}\{\frac{1}{(1-\sqrt{2}t)}+\frac{1}{(1+\sqrt{2}t)}\}$

$\frac{1}{1-t^2}=\frac{1}{(1+t)(1-t)}=\frac{1}{2}[\frac{1}{(1+t)}+\frac{1}{(1-t)}]$

$\therefore \frac{1}{4}\int \frac{1}{2}\{\frac{1}{(1-\sqrt{2}t)}+\frac{1}{1+\sqrt{2}t}\}dt+\frac{1}{4}\int \frac{1}{2}\{\frac{1}{(1-t)}+\frac{1}{(1+t)}\}dt$

$\Rightarrow \frac{1}{8}\{\frac{ln(1-\sqrt{2}t)}{-\sqrt{2}}+\frac{ln(1-t)}{-2}+\frac{ln(1+t)}{2}+c\}$