How have I incorrectly computed $\int \frac{x^{2}+4}{x(x-1)^{2}} dx$? My work:
$\displaystyle \int \frac{x^{2}+4}{x(x-1)^{2}}dx$$ =\int \frac{A}{x} dx+\int \frac{B}{x-1}dx +\int \frac{C}{(x-1)^{2}}dx$
$x^{2} + 4 = A(x-1)^{2} + Bx(x-1) + C(x) $
$x^2 + 4 = x^{2}(A+B) + x(-2A+C) + (A-B)$
equating coefficients:
$1 = A + B $
$0 = - 2A + C $
$4 = A - B$
then
$1 - B = A$
so
$4 = 1 - B - B$
$\frac{-3}{2} = B$
and
$4 = A - \frac{-3}{2}$
$\frac{5}{2} = A$
then
$0 = - 2(\frac{5}{2}) + C$
$0 = - 5 + C$
$5 = C$
plugging A, B, and C back into the integral I got:
$= \frac{5}{2}ln|x| - \frac{3}{2}ln|x - 1| - 5ln|x - 1| + C$
wolfram state this is wrong. What have I done wrong here?
Thank you

$x$ has a $-B$ coefficient as well, so $0 = - 2A - B - C$.  And $4 = A-B$ should just be $4=A$.  It looks like you mistakenly did $Bx(x-1) = Bx^2 - B$ instead of $Bx^2 - Bx$.
Here's a handy tip for dealing with partial fraction decompositions.
When you get to $x^{2} + 4 = A(x-1)^{2} + Bx(x-1) + C(x) $, you can exploit the fact that this is an identity in the variable $x$.  In other words, this equation is true no matter what $x$ is.  So, you can plug in "helpful" values of $x$ to determine the values of $A$, $B$, $C$.
For example, if $x=0$ then $0^2 + 4 = A(-1)^2$, which gives $A = 4$.  Plugging in $x=1$ immediately gives you $C$.  Then since you have $A$ and $C$ you can plug in any other value of $x$ to get $B$.  No need to mess around with systems of equations.
