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

$$\int \frac{1}{(x-1)^2} dx \neq \ln|x-1|+C$$
$$\int \frac{1}{(x-1)^2} dx= \int (x-1)^{-2} dx =\frac{(x-1)^{-1}}{-1}+C=-\frac{1}{x-1}+C$$
