<s>[INST] Partial fractions - integration $$\int \frac{4}{(x)(x^2+4)} $$ 
By comparing coefficients, 
$ 4A = 4 $,
$A = 1$ 
$1 + B = 0 $, 
$B= -1 $ 
$xC= 0 $, 
$C= 0 $ 
where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$. 
So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$. 
And my final answer is 
$\ln|x| - x \ln |x^2 + 4| + C$. 
However my answer is wrong , the answer is - $\ln|x| - \frac{1}{2} \ln |x^2 + 4| + C$. 
Where have I gone wrong?
 [/INST] Yes, $4 = A(x^2+4)~+~(Bx+C)x \implies A=1, B=-1, C=0$
So
$$\require{enclose}\int \dfrac{4}{x(x^2+4)}~\mathrm d x ~{= \int \dfrac{1}{x}+\dfrac{\enclose{circle}{~-x~}}{(x^2+4)}~\mathrm d x \\ = \ln\lvert x\rvert  -\int\dfrac{\tfrac 12\mathrm d (x^2+4)}{(x^2+4)} \\ = \ln\lvert x\rvert -\tfrac 12\ln\lvert x^2+4\rvert+D}$$
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