1. Improper Integral check

hi!

the question and my work are both attached .

Thanks

2. Originally Posted by hoger
hi!

the question and my work are both attached .
$\frac{1}{2x^2-1} \ne \frac{1}{2} \cdot \frac{1}{x^2-1}$
$\frac{1}{2x^2-1} = \frac{1}{(\sqrt{2} x - 1)(\sqrt{2} x + 1)} = \frac{A}{\sqrt{2} x - 1} + \frac{B}{\sqrt{2} x + 1}$