# How do you integrate int 1/(x^2(2x-1)) using partial fractions?

Apr 17, 2018

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

#### Explanation:

We need to find $A , B , C$ such that

$\frac{1}{{x}^{2} \left(2 x - 1\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{2 x - 1}$

for all $x$.

Multiply both sides by ${x}^{2} \left(2 x - 1\right)$ to get

$1 = A x \left(2 x - 1\right) + B \left(2 x - 1\right) + C {x}^{2}$

$1 = 2 A {x}^{2} - A x + 2 B x - B + C {x}^{2}$

$1 = \left(2 A + C\right) {x}^{2} + \left(2 B - A\right) x - B$

Equating coefficients give us

$\left\{\begin{matrix}2 A + C = 0 \\ 2 B - A = 0 \\ - B = 1\end{matrix}\right.$

And thus we have $A = - 2 , B = - 1 , C = 4$. Substituting this in the initial equation, we get

$\frac{1}{{x}^{2} \left(2 x - 1\right)} = \frac{4}{2 x - 1} - \frac{2}{x} - \frac{1}{x} ^ 2$

Now, integrate it term by term
$\int \setminus \frac{4}{2 x - 1} \setminus \mathrm{dx} - \int \setminus \frac{2}{x} \setminus \mathrm{dx} - \int \setminus \frac{1}{x} ^ 2 \setminus \mathrm{dx}$

to get

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

Apr 17, 2018

The answer is $= \frac{1}{x} - 2 \ln \left(| x |\right) + 2 \ln \left(| 2 x - 1 |\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{1}{{x}^{2} \left(2 x - 1\right)} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{2 x - 1}$

$= \frac{A \left(2 x - 1\right) + B x \left(2 x - 1\right) + C \left({x}^{2}\right)}{{x}^{2} \left(2 x - 1\right)}$

The denominators are the same, compare the numerators

$1 = A \left(2 x - 1\right) + B x \left(2 x - 1\right) + C \left({x}^{2}\right)$

Let $x = 0$, $\implies$, $1 = - A$, $\implies$, $A = - 1$

Let $x = \frac{1}{2}$, $\implies$, $1 = \frac{C}{4}$, $\implies$, $C = 4$

Coefficients of ${x}^{2}$

$0 = 2 B + C$

$B = - \frac{C}{2} = - \frac{4}{2} = - 2$

Therefore,

$\frac{1}{{x}^{2} \left(2 x - 1\right)} = - \frac{1}{x} ^ 2 - \frac{2}{x} + \frac{4}{2 x - 1}$

So,

$\int \frac{1 \mathrm{dx}}{{x}^{2} \left(2 x - 1\right)} = - \int \frac{1 \mathrm{dx}}{x} ^ 2 - \int \frac{2 \mathrm{dx}}{x} + \int \frac{4 \mathrm{dx}}{2 x - 1}$

$= \frac{1}{x} - 2 \ln \left(| x |\right) + 2 \ln \left(| 2 x - 1 |\right) + C$