piteraufqvw

2023-02-20

How to integrate $\int \frac{1}{{x}^{2}\left(2x-1\right)}$ using partial fractions?

Peruvianoe4p

We need to find $A,B,C$ such that
$\frac{1}{{x}^{2}\left(2x-1\right)}=\frac{A}{x}+\frac{B}{{x}^{2}}+\frac{C}{2x-1}$
for all $x$.
Multiply both sides by ${x}^{2}\left(2x-1\right)$ to get
$1=Ax\left(2x-1\right)+B\left(2x-1\right)+C{x}^{2}$
$1=2A{x}^{2}-Ax+2Bx-B+C{x}^{2}$
$1=\left(2A+C\right){x}^{2}+\left(2B-A\right)x-B$
Equating coefficients give us
$\left\{\begin{array}{l}2A+C=0\\ 2B-A=0\\ -B=1\end{array}$
And thus we have $A=-2,B=-1,C=4$. When this is substituted in the initial equation, we obtain
$\frac{1}{{x}^{2}\left(2x-1\right)}=\frac{4}{2x-1}-\frac{2}{x}-\frac{1}{{x}^{2}}$
Now, integrate it term by term

to get
$2\mathrm{ln}|2x-1|-2\mathrm{ln}|x|+\frac{1}{x}+C$

davz198888za

Perform the decomposition into partial fractions
$\frac{1}{{x}^{2}\left(2x-1\right)}=\frac{A}{{x}^{2}}+\frac{B}{x}+\frac{C}{2x-1}$
$=\frac{A\left(2x-1\right)+Bx\left(2x-1\right)+C\left({x}^{2}\right)}{{x}^{2}\left(2x-1\right)}$
Compare the numerators because the denominators are the same.
$1=A\left(2x-1\right)+Bx\left(2x-1\right)+C\left({x}^{2}\right)$
Let $x=0$, $⇒$, $1=-A$, $⇒$, $A=-1$
Let $x=\frac{1}{2}$, $⇒$, $1=\frac{C}{4}$, $⇒$, $C=4$
Coefficients of ${x}^{2}$
$0=2B+C$
$B=-\frac{C}{2}=-\frac{4}{2}=-2$
Hence,
$\frac{1}{{x}^{2}\left(2x-1\right)}=-\frac{1}{{x}^{2}}-\frac{2}{x}+\frac{4}{2x-1}$
So,
$\int \frac{1dx}{{x}^{2}\left(2x-1\right)}=-\int \frac{1dx}{{x}^{2}}-\int \frac{2dx}{x}+\int \frac{4dx}{2x-1}$
$=\frac{1}{x}-2\mathrm{ln}\left(|x|\right)+2\mathrm{ln}\left(|2x-1|\right)+C$

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