Partial fractions - integration

Question

$$\int \frac{4}{(x)(x^2+4)} $$

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?

Answer 1

Yes, $4 = A(x^2+4)~+~(Bx+C)x \implies A=1, B=-1, C=0$

So

$$\require{enclose}\int \dfrac{4}{x(x^2+4)}~\mathrm d x ~{= \int \dfrac{1}{x}+\dfrac{\enclose{circle}{~-x~}}{(x^2+4)}~\mathrm d x \\ = \ln\lvert x\rvert -\int\dfrac{\tfrac 12\mathrm d (x^2+4)}{(x^2+4)} \\ = \ln\lvert x\rvert -\tfrac 12\ln\lvert x^2+4\rvert+D}$$

Answer 2

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 $x^2+4$.

Answer 3

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.

Answer 4

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$$\boxed{\int\frac {4\, dx}{x(x^2+4)}=\log x-\tfrac 12\log(x^2+4)+C}$$