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?
 A: Writing 
$$
\frac{4}{x(x^2+4)} = \frac{A}{x} + \frac{Bx+C}{x^2+4},  
$$
you saw that $A=1$, $B=-1$, and $C=0$. 
So you have 
$$
\int \frac{4}{x(x^2+4)}dx  = \int \frac{dx}{x}  - \int \frac{x}{x^2+4} dx. 
$$
To integrate the second term on the right, do a $u$-substitution: 
let $u=x^2+4$. Then $du=2xdx$. 
So 
$$
\int \frac{dx}{x}  - \int \frac{x}{x^2+4} dx 
= \int \frac{dx}{x}  - \frac{1}{2}\int \frac{du}{u}.  
$$
After integrating term-by-term, change the $u$ back into $x^2+4$. 
A: Yes, $4 = A(x^2+4)~+~(Bx+C)x \implies A=1, B=-1, C=0$
So
$$\require{enclose}\int \dfrac{4}{x(x^2+4)}~\mathrm d x ~{= \int \dfrac{1}{x}+\dfrac{\enclose{circle}{~-x~}}{(x^2+4)}~\mathrm d x \\ = \ln\lvert x\rvert  -\int\dfrac{\tfrac 12\mathrm d (x^2+4)}{(x^2+4)} \\ = \ln\lvert x\rvert -\tfrac 12\ln\lvert x^2+4\rvert+D}$$
A: Your partial fraction expansion was correct.$$\frac 4{x(x^2+4)}=\frac 1x-\frac x{x^2+4}$$However, the error lies in integrating the second term of the right-hand side. To simplify$$\int\frac x{x^2+4}\, dx$$Make a substituting $z=x^2+4$. The derivative is $dz=2x\, dx$ so $x\, dx=dz/2$. Therefore, the integral transforms into$$\begin{align*}\int\frac {x\, dx}{x^2+4} & =\frac 12\int\frac {dz}{z}\\ & =\tfrac 12\log z+C\\ & =\tfrac 12\log(x^2+4)+C\end{align*}$$Hence, we have$$\boxed{\int\frac {4\, dx}{x(x^2+4)}=\log x-\tfrac 12\log(x^2+4)+C}$$
A: I'm not necessarily sure how you got a factor of $x$ in the second term, but here is my go at it:
$$\frac{4}{x(x^2+4)}= \frac{A}{x} + \frac{Bx+C}{x^2+4}=\frac{A(x^2+4)+x(Bx+C)}{x(x^2+4)}=\frac{Ax^2+Bx^2+Cx+4A}{x(x^2+4)}$$
Then, by equating the numerator of the first and last expressions we have:
$$4=x^2(A+B) +Cx +4A$$
Equating coefficients:
$$A+B=0,C=0,4A=4\Rightarrow A=1,B=-1 \text{ and } C=0$$
Then we have:
$$\int{\frac{4}{x(x^2+4)}}\,dx=\int{\frac{1}{x}-\frac{x}{x^2+4}\,dx} = \int\frac{dx}{x}\,  - \int{\frac{x}{x^2+4}\,dx}$$
In the second integral, we let $u=x^2+4$ then $du=2x dx\Rightarrow xdx=\frac{1}{2}du$. Hence
$$\int{\frac{dx}{x}} - \frac{1}{2}\int{\frac{du}{u}}=\ln \left|x\right|+\frac{1}{2}\ln\left|u\right|+C=\ln|x|-\frac{1}{2}\ln|x^2+4|+C$$
as required.
