How do you integrate $\int \frac{- 2 x - 3}{x^{2} - x}$ $int (-2x-3)/(x^2-x)$ using partial fractions?

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Answer 1 · 2016-11-17T16:59:19

The answer is $= 3 ln x - 5 ln (x - 1) + C$ $=3lnx-5ln(x-1)+C$

Let's do the decomposition in partial fractions

The denominator $= x^{2} - x = x (x - 1)$ $=x^2-x=x(x-1)$

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

$= \frac{A (x - 1) + B x}{x (x - 1)}$ $=(A(x-1)+Bx)/(x(x-1))$

Therefore,
$- 2 x - 3 = A (x - 1) + B x$ $-2x-3=A(x-1)+Bx$

let $x = 1$ $x=1$ , $\Rightarrow$ $=>$ $- 5 = B$ $-5=B$

let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ $- 3 = - A$ $-3=-A$ $\Rightarrow$ $=>$ $A = 3$ $A=3$

So, $\frac{- 2 x - 3}{x (x - 1)} = \frac{3}{x} - \frac{5}{x - 1}$ $(-2x-3)/(x(x-1))=3/x-5/(x-1)$

$\int \frac{(- 2 x - 3) d x}{x (x - 1)} = \int \frac{3 d x}{x} - \int \frac{5 d x}{x - 1}$ $int((-2x-3)dx)/(x(x-1))=int(3dx)/x-int(5dx)/(x-1)$

$= 3 ln x - 5 ln (x - 1) + C$ $=3lnx-5ln(x-1)+C$

as $\int \frac{d x}{x} = ln x + c$ $intdx/x=lnx+c$

How do you integrate ∫−2x−3x2−xint (-2x-3)/(x^2-x) using partial fractions?