# How do you integrate int (3x-1)/(x^2-x) using partial fractions?

Apr 12, 2018

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

#### Explanation:

Perform the decomposition into partial fractions

$\frac{3 x - 1}{{x}^{2} - x} = \frac{3 x - 1}{x \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x - 1}$

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

The denominators are the same, compare the numerators

$3 x - 1 = A \left(x - 1\right) + B x$

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

Let $x = 1$, $\implies$, $2 = B$, $\implies$, $B = 2$

Therefore,

$\frac{3 x - 1}{{x}^{2} - x} = \frac{1}{x} + \frac{2}{x - 1}$

$\int \frac{\left(3 x - 1\right) \mathrm{dx}}{{x}^{2} - x} = \int \frac{1 \mathrm{dx}}{x} + \int \frac{2 \mathrm{dx}}{x - 1}$

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

Apr 12, 2018

$\ln | x | + 2 \ln | \left(x - 1\right) | + C , \mathmr{and} , \ln | x {\left(x - 1\right)}^{2} | + C$.

#### Explanation:

Suppose that, $I = \int \frac{3 x - 1}{{x}^{2} - x} \mathrm{dx} = \int \frac{3 x - 1}{x \left(x - 1\right)} \mathrm{dx}$

To decompose the integrand $\frac{3 x - 1}{x \left(x - 1\right)}$ into partial

fraction, we let, for some $A , B \in \mathbb{R}$,

$\frac{3 x - 1}{x \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x - 1} = \frac{A \left(x - 1\right) + B x}{x \left(x - 1\right)}$,

$\Rightarrow A \left(x - 1\right) + B x = 3 x - 1. \ldots \ldots \ldots . \left(\diamond\right)$.

$\text{Note that "(diamond)" must hold "AA x in RR," in particular so, for}$

$x = 0 , \mathmr{and} x = 1$.

 x=0, x=1, and (diamond) rArr A=1, &, B=2," resp.".

$\therefore I = \int \left\{\frac{1}{x} + \frac{2}{x - 1}\right\} \mathrm{dx}$,

$= \ln | x | + 2 \ln | \left(x - 1\right) |$.

$\Rightarrow I = \ln | x | + 2 \ln | \left(x - 1\right) | + C , \mathmr{and} , \ln | x {\left(x - 1\right)}^{2} | + C$.

Enjoy Maths.!

Apr 12, 2018

$2 \ln | \left(x - 1\right) | + \ln | x | + C$.

#### Explanation:

The Question can easily be solved without using the

Method of Partial Fraction, as shown below :

$\int \frac{3 x - 1}{{x}^{2} - x} \mathrm{dx} = \int \frac{3 x - 1}{x \left(x - 1\right)} \mathrm{dx}$,

$= \int \left\{\frac{3 x}{x \left(x - 1\right)} - \frac{1}{x \left(x - 1\right)}\right\} \mathrm{dx}$,

$= \int \left\{\frac{3}{x - 1} - \frac{x - \left(x - 1\right)}{x \left(x - 1\right)}\right\} \mathrm{dx}$,

$= \int \left[\frac{3}{x - 1} - \left\{\frac{x}{x \left(x - 1\right)} - \frac{x - 1}{x \left(x - 1\right)}\right\}\right] \mathrm{dx}$,

$= 3 \int \frac{\mathrm{dx}}{x - 1} - \int \frac{\mathrm{dx}}{x - 1} + \int \frac{\mathrm{dx}}{x}$,

$= 2 \int \frac{\mathrm{dx}}{x - 1} + \int \frac{\mathrm{dx}}{x}$,

$= 2 \ln | \left(x - 1\right) | + \ln | x | + C$, as before!

Enjoy Maths!