Express in partial fractions:
a) (3x + 1) / (x + 1)^2
b) (x^2 + 7x + 2) / (1 + x^2)(2 - x)
a)
$\displaystyle \frac {3x + 1} {(x + 1)^2} $ (it's the repeated factor type) so
$\displaystyle \frac {3x + 1} {(x + 1)^2}=\frac {A}{x+1}+\frac {B}{(x+1)^2}$
$\displaystyle \frac {3x + 1} {(x + 1)^2}=\frac {A(x+1)+B}{(x+1)^2}$
you can cancel the denominator,
so you have
$\displaystyle 3x + 1=A(x+1)+B$
you just have to find the value for A and B and then change A and B from this eq...$\displaystyle \frac {A}{x+1}+\frac {B}{(x+1)^2}$
....hope it help.....